how to reverse string in array in C

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 string
    for (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

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 0
for(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 0
for(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 0
    for(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 0
for(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 0
for(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 0
    for(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 0
    for(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 52
ABCDEFGHIJKLMNOPQRSTUVWXYZZYXWVUTSRQPONMLKJIHGFEDCBA

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