在C中的for循环随机中断。

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英文:

For loop in C breaks randomly

问题

我一直在处理这个C循环有一段时间了。我试图通过for循环创建一个字符串数组（我不确定我是否做得正确。我希望我是）。每当我输入一个带有空格的字符串时，for循环就会中断并跳过所有迭代。例如，如果我在命令行中输入S 1，它会中断。

这是代码：

  1. #include <stdlib.h>
  2. #include <stdio.h>
  3. #include <math.h>
  4. #include <string.h>
  6. int main() {
  7.     int players;
  8.     int jerseys;
  9.     int count = 0;
  10.     int i;
  11.     scanf("%d", &jerseys);
  13.     scanf("%d", &players);
  15.     char size[jerseys], p[players][100];
  17.     for (jerseys; jerseys > 0; jerseys--) {
  18.         scanf(" %c", &size[count]);
  19.         count++;
  20.     }
  21.     getchar();
  22.     count = 0;
  23.     for (players; players > 0; players--) {
  24.         /*scanf(" %s", p[0] );    */    /*你不能在C中分配数组。*/
  25.         getchar();
  26.         fgets(p[count], 100, stdin);
  27.         printf("%s", p[count]);
  28.         printf("%c", p[count][2]);  /* 第29行 */
  29.         printf("Hello\n");
  30.         count++;
  31.     }
  33.     return 0;
  34. }

此外，在第29行，如果我将索引从2更改为1，循环会立即中断，无论我放入什么。

我有一个Python代码，从C中本质上想要的：

  1. given = []
  2. jerseys = int(input())
  3. if jerseys == 0:
  4.     print(0)
  5. players = int(input())
  6. j = []
  7. requests = 0
  8. for _ in range(jerseys):
  9.     size = input()
  10.     j.append(size)
  11. for _ in range(players):
  12.     p = input().split()

我已经查看了很多地方，我认为问题出在数组上，而不是换行符，但我一点头绪。

编辑：

这是看起来像我想要输入的内容（也是我通常尝试的内容）：

  1. 3
  2. 3
  3. S
  4. M
  5. L
  6. S 1
  7. S 3
  8. L 2
英文:

I've been struggling with this loop in C for a while now. I'm trying to create a string array through a for loop (which I'm not sure I'm doing correctly. I hope I am). Every time I enter a string with a space in it, the for loop breaks and skips all iterations. For example, if I write S 1 in the command line, it would break.

This is the code:

  1. #include &lt;stdlib.h&gt;
  2. #include &lt;stdio.h&gt;
  3. #include &lt;math.h&gt;
  4. #include &lt;string.h&gt;
  6. int main(){
  7. 	int players;
  8. 	int jerseys;
  9. 	int count = 0;
  10. 	int i;
  11. 	scanf(&quot;%d&quot;, &amp;jerseys);
  12. 	
  13. 	scanf(&quot;%d&quot;, &amp;players);
  14. 	
  15. 	char size[jerseys], p[players][100];
  17. 	for(jerseys; jerseys &gt; 0; jerseys--){
  18. 		
  19. 		scanf(&quot; %c&quot;, &amp;size[count]); 
  20. 		count++;
  21. 	}
  22. 	getchar();
  23. 	count = 0;
  24. 	for(players; players&gt;0; players--){
  25. 		/*scanf(&quot; %s&quot;, p[0] );	*/	/*you cant assign arrays in C.*/
  26. 		getchar();
  27. 		fgets(p[count], 100, stdin);
  28. 		printf(&quot;%s&quot;, p[count]);
  29. 		printf(&quot;%s&quot;, p[count][2]);  /* LINE 29 */
  30. 		printf(&quot;Hello\n&quot;);
  31. 		count ++;
  32. 	}		
  35. 	return 0;
  36. }

Moreover, on line 29, if I change the index from 2 to 1, the loop instantly breaks, no matter what I put.

I have a python code for what I essentially want from C:

  1. given = []
  2. jerseys = int(input())
  3. if jerseys == 0:
  4. 	print(0)
  5. players = int(input())
  6. j = []
  7. requests = 0
  8. for _ in range(jerseys):
  9. 	size = input()
  10. 	j.append(size)
  11. for _ in range(players):
  12. 	 p = input().split()

I've looked at many places, and I think the problem is with the array, not the new lines, but I have no clue.

Edit:

This is something that would look like what I want to input(and what I usually try):

  1. 3
  2. 3
  3. S
  4. M
  5. L
  6. S 1
  7. S 3
  8. L 2

答案1

得分: 1

> 如果输入的字符与控制字符不匹配或者与格式化输入的类型不符，那么scanf将终止，留下有问题的字符作为下一个要读取的字符。

如果你在命令行中输入*1\，那么jerseys会被设置为1，但players会是一个随机整数，因为不匹配%d的格式。所以在你的程序中，players`变量可能是一个很大的整数。

所以当你使用scanf时，最好检查返回值，如下所示：

  1. if (scanf("%d", &players) != 1) {
  2.     /* 处理错误 */
  3. }

我运行代码时出现了分段错误。

英文:

> If the input characters do not match the control characters or are of the wrong type for a formatted input scanf terminates leaving the offending character as the next character to be read.

If you write 1` in the command line, then jerseys set to 1, but players is a random int because the ` not match the %d format. So in you program, your players variable may be a big int.

So when you use scanf, you'd better to check the return value like

  1. if ((scanf(&quot;%d&quot;, &amp;players) != 1) {
  2.     /* error handle */
  3. }

I run the code and segmentation fault is raise.

答案2

得分: 1

发布的代码无法干净地编译！

以下是来自 gcc 编译器的输出：

  1. gcc -ggdb3 -Wall -Wextra -Wconversion -pedantic -std=gnu11 -c "untitled.c" -o "untitled.o"
  3. untitled.c: 在函数‘main’中：
  4. untitled.c:17:5: 警告：没有效果的语句 [-Wunused-value]
  5.    17 |     for(jerseys; jerseys > 0; jerseys--){
  6.       |     ^~~
  7.       
  8. untitled.c:24:5: 警告：没有效果的语句 [-Wunused-value]
  9.    24 |     for(players; players>0; players--){
  10.       |     ^~~
  11.       
  12. untitled.c:29:18: 警告：格式‘%s’预期的参数类型是‘char *’，但参数2的类型是‘int’ [-Wformat=]
  13.    29 |         printf("%s", p[count][2]);  /* LINE 29 */
  14.       |                 ~^   ~~~~~~~~~~~
  15.       |                  |           |
  16.       |                  char *      int
  17.       |                 %d
  18.       
  19. untitled.c:10:9: 警告：未使用的变量‘i’ [-Wunused-variable]
  20.    10 |     int i;
  21.       |         ^
  23. 编译成功完成。

请更正代码并检查C库I/O函数返回的状态。

关于：

  1. 编译成功完成。

由于编译器输出了几个警告，这个语句仅表示编译器对你的意图进行了一些（不一定正确的）猜测。

英文:

the posted code does not cleanly compile!

Here is the output from the gcc compiler:

  1. gcc -ggdb3 -Wall -Wextra -Wconversion -pedantic -std=gnu11 -c &quot;untitled.c&quot; -o &quot;untitled.o&quot;
  3. untitled.c: In function ‘main’:
  4. untitled.c:17:5: warning: statement with no effect [-Wunused-value]
  5.    17 |     for(jerseys; jerseys &gt; 0; jerseys--){
  6. 	  |     ^~~
  7. 	  
  8. untitled.c:24:5: warning: statement with no effect [-Wunused-value]
  9.    24 |     for(players; players&gt;0; players--){
  10. 	  |     ^~~
  11. 	  
  12. untitled.c:29:18: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat=]
  13.    29 |         printf(&quot;%s&quot;, p[count][2]);  /* LINE 29 */
  14. 	  |                 ~^   ~~~~~~~~~~~
  15. 	  |                  |           |
  16. 	  |                  char *      int
  17. 	  |                 %d
  18. 	  
  19. untitled.c:10:9: warning: unused variable ‘i’ [-Wunused-variable]
  20.    10 |     int i;
  21. 	  |         ^
  22. 	  
  23. Compilation finished successfully.

Please correct the code AND check the returned status from the C library I/O functions

Regarding:

  1. Compilation finished successfully.

since the compiler output several warnings, This statement only means the compiler made some (not necessarily correct) guesses as to what you meant.

huangapple
  • 本文由 发表于 2023年2月8日 13:22:05
  • 转载请务必保留本文链接：https://go.coder-hub.com/75381645.html
  • arrays
  • c
  • for-loop
  • python
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