我的strcpy有什么问题

我的strcpy有什么问题

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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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07-29 19:58