C语言的问题（新手提问）

[xadm1986]

帮忙看一下这个程序问题出在哪？
#include "stdio.h"
int main()
{
char *a = "abcd";
char *b = "1234";
strcpy1(a, b);
return 0;
}
void strcpy1(char *dest, char *src)
{
dest = src;
}

编译的时候报错：
"strcpy1.c", line 9.6: 1506-343 (S) Redeclaration of strcpy1 differs from previous declaration on li
ne 6 of "strcpy1.c".
"strcpy1.c", line 9.6: 1506-050 (I) Return type "void" in redeclaration is not compatible with the p
revious return type "int".

[wade_2003]

#include "stdio.h"
void strcpy1(char *dest, char *src)
{
 dest = src;
}


int main()
{
 char *a = "abcd";
 char *b = "1234";
 strcpy1(a, b);
 return 0;
}

[wade_2003]

函数要先有声明，才能用的，要不就将函数放到前面去

[lirg8405]

void strcpy1(char *dest, char *src)
{
dest = src;
}
把这个移到前面，不会报错，但是你这个函数不能实现你想要的功能

[冻结]

#include "stdio.h"
int main()
{
 char *a = "abcd"; //  要用数组char a[xx];
 char *b = "1234";
 strcpy1(a, b);
 return 0;
}
void strcpy1(char *dest, char *src)
{
 dest = src;  // 要按字符依次拷贝。
}

[huanmie_09]

#include "stdio.h"

char *strcpy1(char *dest, char *src);  //先进行声明

int main()
{
 char *a = "abcd";
 char *b = "1234";
 a = strcpy1(a, b);  //通过返回值的形式获取指针a
 puts(a);
 return 0;
}
char *strcpy1(char *dest, char *src) //返回值类型改一下
{
 dest = src;
 return dest;
}

[vanchristin]

#include "stdio.h"
void strcpy1(char *dest, char *src);
int main()
{
 char *a = "abcd";
 char *b = "1234";
 strcpy1(a, b);
 return 0;
}
void strcpy1(char *dest, char *src)
{
 dest = src;
}

不过这样还是达不到你想要的字符串复制的效果

[killernan]

[Quote=引用 6 楼 vanchristin 的回复:]
C/C++ code

#include "stdio.h"
void strcpy1(char *dest, char *src);
int main()
{
char *a = "abcd";
char *b = "1234";
strcpy1(a, b);
return 0;
}
void strcpy1(char *dest, char *src)
{
……
[/Quote]
strcpy就这么简单？？？？？如果就这么一句就能实现还有必要定义一个库函数吗

[z569362161]

#include "stdio.h"
void strcpy1(char **dest, char **src)
{
    *dest =*src;
}


int main()
{
    char *a = "abcd";
    char *b = "1234";
    strcpy1(&a,&b);
    puts(a);
    return 0;
}

通过2级指针

[z569362161]

strcpy就这么简单？？？？？如果就这么一句就能实现还有必要定义一个库函数吗

就是这么简单。

[xadm1986]

谢谢大家~~~懂了