我的strcpy有什么问题？

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我试图让 strcpy 自己。它应该工作，我甚至复制和粘贴（几乎完全代码）从某人在这里关于 strcpy 。两者都给我一个分段错误。

char * strcpy（char * destination，const char * source） { while（（* destination ++ = * source ++）！='\0'）; return destination; }

此代码有什么问题？

char * a =hello cout< strcpy（a，Haha）<< endl;

解决方案

因此，当你调用 strcpy char a [] =hello; cout< strcpy（a，Haha）<< endl;

strcpy 函数，复制后，目标将指向字符串的结尾，需要返回字符串的开头

I tried to make strcpy myself. It should work, I even copied and pasted the (almost exact code) from someones post here about strcpy. Both give me a "Segmentation Fault".

char* strcpy(char * destination, const char * source) { while( (*destination++ = *source++) != '\0' ) ; return destination; }

What's wrong with this code?

char* a = "hello"; cout << strcpy(a, "Haha") << endl;

解决方案

You are trying to write to data segment since "hello" is stored there.

Therefore, when you call strcpy you get segmentation fault.

Try:

char a[] = "hello"; cout << strcpy(a, "Haha") << endl;

instead.

EDIT: Inside your strcpy function, after the copy, destination will point to end of the string, you need to return beginning of the string instead.