传递字符串作为参数通过的char * [英] Pass string literal as char* via argument

问题描述

我做了哪些修改字符串的函数，请参阅以下code。

I made a function which change string, see the following code.

void Test(char* str, char c) {
    str[1] = c;
}

int main(){
    Test("Hi", '2');
}

我注意到它做了一些运行时错误。我知道如何prevent错误。

I notice it made some run time error. I know how to prevent the error.

char buff[3] = "Hi";
Test(buff,'2');

但我不知道为什么第一个例子做运行时错误。我想，如果我直接传递字符串，就变成为const char。没有人解释究竟发生什么事呢？

but I don't know why the first example made run time error. I guess, if I pass string directly, it becomes const char. Does anyone explain what happened exactly?

PS。
如果我用的char *海峡=然后将它传递到参数？

ps. what if I use char* str = "hi", then pass it into the argument?

char* buff = "Hi";
Test(buff,'2');

这样。我可以修改BUFF？

like this. Can I modify buff?

推荐答案

由于嗨是文字的，这是不允许的字符串是修改，它们是只读的（字符串类型为为const char [N] ）。

Because "Hi" is string literal and it's not allowed to be modified, they are read-only (the type of string literal is const char[n]).

修改它的未定义行为的

关于您的编辑：的char *海峡=喜是无效的，应该是为const char *海峡=喜。这是指向为const char 。再次，修改它被禁止。

Regarding your edit: char* str = "hi" is invalid, it should be const char* str = "hi". Which is pointer to const char. Again, modifying it is disallowed.

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