如何使用指针更改字符串中的字符? [英] How to change a character in a string using pointers?

问题描述

我对此代码有疑问

int main() {
     char *My_St = "abcdef";
     *(My_St+1)='+';
     printf("%s\n",My_St);
     return 0;
}

我构建了这段代码，没有错误，但是当我尝试运行它时，它抛出了分段错误，有人可以告诉我什么地方出问题了

i built this code and has no errors, but when i try to run it, it throws a segmentation fault, could someone tell what's wrong

推荐答案

不能，因为试图修改const数据.

You can't because you are trying to modify const data.

将其更改为:

char My_St[] = "abcdef";

然后您将能够对其进行更改.

Then you will be able to change it.

想想您在做什么，您在声明指向"abcdef"的指针.它是一个指针，不是字符数组.我的意思是"abcdef"生活在服务器场的.text区域中，而这是不可变的.

Think about what you were doing, you were declaring a pointer that pointed to "abcdef". It IS a pointer, not an array of chars. "abcdef" lives in the farm, I mean, in the .text area of your program and that is immutable.

当按照我已说明的方式进行操作时，您正在告诉编译器:我声明此数组，该数组将具有容纳"abcdef"所需的所有字符，并在您存在的地方复制"abcdef".

When you do it the way I've shown, you are telling the compiler: i'm declaring this array, that will have as many chars as are needed to accommodate "abcdef" and also, as you are there, copy "abcdef" to it.

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