为什么这段代码修改一个字符串不工作？ [英] Why does this code to modify a string not work?

问题描述

对于c风格字符串，如何将字符指定给字符指针指向的内存地址？例如，在下面的示例中，我想将num更改为123456，因此我尝试将p设置为0所在的数字，并尝试用4覆盖它。非常感谢。

With c-style strings, how do you assign a char to a memory address that a character pointer points to? For example, in the example below, I want to change num to "123456", so I tried to set p to the digit where '0' is located and I try to overwrite it with '4'. Thanks.

#include <stdio.h>
#include <stdlib.h>

int main()
{
    char* num = (char*)malloc(100);
    char* p = num;

    num = "123056";

    p = p+3;    //set pointer to where '4' should be
    p = '4';

    printf("%s\n", num );

    return 0;
}

推荐答案

，因为行：

num = "123056";

更改 num 内存（和 p 仍然指向那个内存，所以它们不再是同一个位置）到最可能的只读内存。

changes num to point away from the allocated memory (and p remains pointing to the that memory so they're no longer the same location) to what is most likely read-only memory. You are not permitted to change the memory belonging to string literals, it's undefined behaviour.

您需要以下内容：

#include <stdio.h>
#include <stdlib.h>
int main (void) {
    char *num = malloc (100); // do not cast malloc return value.
    char *p = num;

    strcpy (num, "123056");   // populate existing block with string.

    p = p + 3;                // set pointer to where '0' is.
    *p = '4';                 // and change it to '4'.

    printf ("%s\n", num );    // output it.

    return 0;
}

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