我的strcpy有什么问题？ [英] What's wrong with my strcpy?
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问题描述
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.
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