替换字符串 C 的单个字符元素

发布于 2025-01-02 19:48:14 字数 545 浏览 0 评论 0原文

我正在尝试在 C 上做一些非常基本的事情，但我不断遇到分段错误。我想做的就是用不同的字母替换一个单词的字母 - 在这个例子中用 L 替换 l。任何人都可以帮助解释我哪里出了问题吗？我认为这应该是一个非常基本的问题，我只是不知道为什么它不起作用。

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

int main(int argc, char *argv[])
{
    char *string1;

    string1 = "hello";
    printf("string1 %s\n", string1);    

    printf("string1[2] %c\n", string1[2]);
    string1[2] = 'L';
    printf("string1 %s\n", string1);

    return 0;
}

对于我的输出我得到

字符串1你好
字符串1[2] l
分段错误

谢谢！

原文

I am trying to do something really basic on C but I keep getting a segmentation fault. All I want to do is replace a letter of a word with a different letter- in this example replace the l with a L. Can anyone help explain where I have gone wrong? It should be a really basic problem I think, I just have no idea why its not working.

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

int main(int argc, char *argv[])
{
    char *string1;

    string1 = "hello";
    printf("string1 %s\n", string1);    

    printf("string1[2] %c\n", string1[2]);
    string1[2] = 'L';
    printf("string1 %s\n", string1);

    return 0;
}

For my output I get

string1 hello
string1[2] l
Segmentation fault

Thanks!

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[旋木] 2025-01-09 19:48:14

string1 = "你好";
字符串1[2] = 'L';

您无法更改字符串文字，这是未定义的行为。试试这个：

char string1[] = "hello";

或者也许：

char *string1;
string1 = malloc(6); /* hello + 0-terminator */
strcpy(string1, "hello");

/* Stuff. */

free(string1);

string1 = "hello";
string1[2] = 'L';

You can't change string literals, it's undefined behavior. Try this:

char string1[] = "hello";

Or maybe:

char *string1;
string1 = malloc(6); /* hello + 0-terminator */
strcpy(string1, "hello");

/* Stuff. */

free(string1);

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节枝 2025-01-09 19:48:14

char *string1;
string1 = "hello";

string1 指向字符串文字，并且字符串文字是不可修改的。

您可以做的是使用字符串文字的元素初始化数组。

char string1[] =  "hello";

string1 数组的元素是可以修改的。

char *string1;
string1 = "hello";

string1 points to a string literal and string literals are non-modifiable.

What you can do is initialize an array with the elements of a string literal.

char string1[] =  "hello";

the elements of string1 array are modifiable.

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め可乐爱微笑 2025-01-09 19:48:14

char *string1 = "hello";

运行代码时，字符串文字将位于只读部分中。操作系统不允许代码更改该内存块，因此您会遇到段错误。

char string1[] = "hello";

当您运行代码时，字符串文字将被推入堆栈。

char *string1 = "hello";

When running the code, the string literal will be in a section that is read-only. OS does not allow the code to change that block of memory, so you get a seg-fault.

char string1[] = "hello";

The string literal will be pushed onto the stack when you run the code.

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未蓝澄海的烟 2025-01-09 19:48:14

 string1[2] = 'L';

您正在尝试更改字符串文字，这在 C 中是未定义的行为。
而是使用 string1[]="hello";
您遇到的分段错误是因为文字可能存储在内存的只读部分中，并且尝试写入它会产生未定义的行为。

 string1[2] = 'L';

you are trying to change a string literal which is undefined behavior in C.
Instead use string1[]="hello";
Segmentation fault you get is because the literal is probably stored in the the read only section of the memory and trying to write to it produces undefined behavior.

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