拷贝字符串出现的问题 - C语言论坛

已结贴√ 问题点数：20 回复次数：2

拷贝字符串出现的问题

#include<stdio.h>
#include<string.h>
main()
{
    static char *s="Chinese";
    static char *t="American";
    printf("\n\ns=%s",s);
    printf("\nt=%s",t);
    strcpy(s,t);
         printf("Copy After:\n");
    printf("\ns=%s",s);
    printf("\nt=%t",t);
}
结果为：
s=Chinese
t=American
Copy After:
s=American
t=
为什么拷贝完字符串后t的值会丢失？如果不定义成常量，用scanf函数读入两个相同的字符串，再进行复制拷贝，就不会出现字符串丢失现象。什么地方出错了？请高手指教。

#include<stdio.h>
#include<string.h>
main()
{
    char *s,*t;
         scanf("%s%s",s,t);
    printf("\n\ns=%s",s);
    printf("\nt=%s",t);
    strcpy(s,t);
         printf("Copy After:\n");
    printf("\ns=%s",s);
    printf("\nt=%t",t);
}
结果为：
s=Chinese
t=American
Copy After:
s=American
t=American

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第 2 楼

得分:10

C99标准   第130页

  32   EXAMPLE   8   The   declaration
  char   s[]   =   "abc",   t[3]   =   "abc";

  defines   "plain"   char   array   objects   s   and   t   whose   elements   are   initialized   with   character   string
  literals.

  This   declaration   is   identical   to
  char   s[]   =   {   'a',   'b',   'c',   '\0'   },
  t[]   =   {   'a',   'b',   'c'   };
  The   contents   of   the   arrays   are   modifiable.

  On   the   other   hand,   the   declaration
  char   *p   =   "abc";
  defines   p   with   type   "pointer   to   char"   and   initializes   it   to   point   to   an   object   with   type
  "array   of   char"   with   length   4   whose   elements   are   initialized   with   a   character   string   literal.
  If   an   attempt   is   made   to   use   p   to   modify   the   contents   of   the   array,   the   behavior   is   undefined.

这个要看编译器的，
一般情况下我是这么理解的，char *p中p指向常量数据段，且p是可变的，它指向的内容的修改要看你的编译器是否支持了。char p[]中p指向局部数据段

你第一个例子最好改成字符数组形式。

第 3 楼

得分:10

这样看看就清楚了

main()
{
    static char s[]="Chinese";
    static char *t="American";
    char* p=t;
    printf("\ns=%s",s);
    printf("\nt address:%p t=%s",t,t);
    strcpy(s,t);
         printf("\nCopy After:\n");
    printf("\ns=%s",s);
    printf("\nt address:%p t=%s",t,t);
    printf("\np address:%p p=%s",p,p);
}