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- #include <stdio.h>;
- char *
- find_char(char const *source,char const *chars)
- {
- char a;
- int i,j;
- if ( (source == NULL) || (chars == NULL) )
- return NULL;
- j=0;
- while (*(source+j) != '\0'){
- i=0;
- while (*(chars+i)!='\0'){
- if (*(chars+i) == *(source+j))
- return (source+j);
- i++;
- }
- j++;
- }
- return NULL;
- }
- int
- main()
- {
- char a[]="fsabdfsf";
- char b[]="jhkhjkhjb";
- char *c;
- c=find_char(a,b);
- printf("%s\n",c);
- }
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这段程序可以编译,执行暂时也正常,但在编译时一个warning,大家帮我看看应该怎样改正.
- [root@aphhk point]# cc 6.18.1.c
- 6.18.1.c: In function `find_char':
- 6.18.1.c:15: warning: return discards qualifiers from pointer target type
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