英文:
how to reverse string in array in C
问题
我已翻译以下代码段：
char *problem58(){char temp, *result = (char *)malloc(52 * sizeof(char));int left = 0, right = 52 - 1;for (char i = 'A'; i <= 'Z'; i++){result[left] = i;printf("%c", i);left++;}// 这是我用来反转字符串的方法for (int i = left; i < right; i++){temp = result[left];result[left] = result[right];result[right] = temp;printf("%c", result[right]);right--;}return result;}
运行此行代码在终端中输出：ABCDEFGHIJKLMNOPQRSTUVWXYZSW\:C=cepSmoC
英文:
I have exercise to use for loop to initialization 1 string array from A to Z and then Z to A.
I was successful to print from A to Z but print Z to A not working. So i need to help in this code.
char *problem58(){char temp, *result = (char *)malloc(52 * sizeof(char));int left = 0, right = 52 - 1;for (char i = 'A'; i <= 'Z'; i++){result[left] = i;printf("%c", i);left++;}// this is my idea to reverse this stringfor (int i = left; i < right; i++){temp = result[left];result[left] = result[right];result[right] = temp;printf("%c", result[right]);right--;}return result;}
line code run in terminal: ABCDEFGHIJKLMNOPQRSTUVWXYZSW\:C=cepSmoC
答案1
得分: 0
这是我的答案
#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '\0'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('\0')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '\0'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('\0')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '\0'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('\0')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '\0'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('\0')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}
运行它会输出
str's length is 52ABCDEFGHIJKLMNOPQRSTUVWXYZZYXWVUTSRQPONMLKJIHGFEDCBA
英文:
Here is my answer
#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '\0'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('\0')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '\0'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('\0')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '\0'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('\0')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('#include <stdio.h>#include <string.h>#include <stdlib.h>char* problem58(){int n = 0;int left = 0, right = 52 - 1;// A--->Z and Z--->A total number is 52, because string end with '\0'(0), so add 1 is 53.char* result = (char*)malloc(53 * sizeof(char));// Check if it is possible to dynamically allocate memory for 53 variables of type "char"// You can also write like this: if(result == NULL)if(NULL == result){printf("No space\n");exit(1);}// Initial result with 0,because string end with ('\0')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}')0.// So result[0] will be 0, result[1] will be 0 ...... result[52] will be 0for(int m = 0; m < 53;m++){result[m] = 0;}for (char i = 'A'; i <= 'Z'; i++){result[left] = i;//printf("%c", i);left++;}// Now left is 26, reset it to 0, because result[0] is 'A'left = 0;// n is execute times, need to let result[51] = result[0] = 'A', and result[50] = result[1] = 'B' ......//so right-- and left++for (n = 0; n < 26; n++){result[right] = result[left];right--;left++;}return result;}int main(){char* str = problem58();printf("str's length is %ld\n",strlen(str));printf("%s\n",str);free(str);return 0;}
Run it will output
str's length is 52ABCDEFGHIJKLMNOPQRSTUVWXYZZYXWVUTSRQPONMLKJIHGFEDCBA
答案2
得分: -1
这个for循环
for (int i = left; i < right; i++){temp = result[left];result[left] = result[right];//..
由于动态分配的数组未初始化，因此调用未定义行为，结果是表达式 result[left] 和 result[right] 具有不确定的值。数组中没有要交换的内容。数组必须填充为字符。
而且，无论如何，该数组都不包含字符串，因为它没有以终止零字符 '\0' 结尾。
如果我正确理解了你的任务，你需要类似以下的代码。请注意，该函数可以以不同的方式实现。
例如，
char * problem58( void ){char *s = malloc( 2 * ('Z' - 'A' + 1) + 1 );if ( s != NULL ){char *p = s;char c = 'A';do{*p++ = c;} while ( c++ != 'Z' );while ( c != 'A' ){*p++ = --c;}*p = 'char * problem58( void ){char *s = malloc( 2 * ('Z' - 'A' + 1) + 1 );if ( s != NULL ){char *p = s;char c = 'A';do{*p++ = c;} while ( c++ != 'Z' );while ( c != 'A' ){*p++ = --c;}*p = '\0';}return s;}';}return s;}
或者
char * problem58( void ){size_t n = 2 * ('Z' - 'A' + 1);char *s = malloc( n + 1 );if (s != NULL){char *first = s, *last = s + n;*last = 'char * problem58( void ){size_t n = 2 * ('Z' - 'A' + 1);char *s = malloc( n + 1 );if (s != NULL){char *first = s, *last = s + n;*last = '\0';char c = 'A';do{*first++ = *--last = c;} while (c++ != 'Z');}return s;}';char c = 'A';do{*first++ = *--last = c;} while (c++ != 'Z');}return s;}
然后在主函数中调用该函数，如下所示：
char *s = problem58();if ( s != NULL ) puts( s );free( s );
调用该函数的预期输出是：
ABCDEFGHIJKLMNOPQRSTUVWXYZZYXWVUTSRQPONMLKJIHGFEDCBA
英文:
This for loop
for (int i = left; i < right; i++){temp = result[left];result[left] = result[right];//..
invokes undefined behavior because the dynamically allocated array was not initialized and as a result the expressions result[left] and result[right] have indeterminate values. There is nothing to swap in the array. The array has to be filled with characters.
Also the array in any case does not contain a string because it is not appended with the terminating zero character '\0';.
If I have understood the assignment correctly then you need something like the following. Pay attention to that the function can be implemented in different ways.
For example
char * problem58( void ){char *s = malloc( 2 * ( 'Z' - 'A' + 1 ) + 1 );if ( s != NULL ){char *p = s;char c = 'A';do{*p++ = c;} while ( c++ != 'Z' );while ( c != 'A' ){*p++ = --c;}*p = 'char * problem58( void ){char *s = malloc( 2 * ( 'Z' - 'A' + 1 ) + 1 );if ( s != NULL ){char *p = s;char c = 'A';do{*p++ = c;} while ( c++ != 'Z' );while ( c != 'A' ){*p++ = --c;}*p = '\0';}return s;}';}return s;}
or
char * problem58( void ){size_t n = 2 * ( 'Z' - 'A' + 1 );char *s = malloc( n + 1 );if (s != NULL){char *first = s, *last = s + n;*last = 'char * problem58( void ){size_t n = 2 * ( 'Z' - 'A' + 1 );char *s = malloc( n + 1 );if (s != NULL){char *first = s, *last = s + n;*last = '\0';char c = 'A';do{*first++ = *--last = c;} while (c++ != 'Z');}return s;}';char c = 'A';do{*first++ = *--last = c;} while (c++ != 'Z');}return s;}
And the function is called in main like
char *s = problem58();if ( s != NULL ) puts( s );free( s );
The expected output of a call of the function is
ABCDEFGHIJKLMNOPQRSTUVWXYZZYXWVUTSRQPONMLKJIHGFEDCBA
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论