c - 警告 : assignment makes integer from pointer without a cast with chars and strings

Question

我在这段代码上不断遇到段错误，我现在正在学习 C..有人可以帮助我吗？

错误位置:

char *m;
char *as = concat("helloworld", buf);
*m = sha1(as);                               <<<<<< as
printf("%s\n", m);
free(as);

concat 函数(不是我的，用于连接 2 个字符串):

char* concat(char *s1, char *s2)
{
    size_t len1 = strlen(s1);
    size_t len2 = strlen(s2);
    char *result = malloc(len1+len2+1);//+1 for the zero-terminator
    //in real code you would check for errors in malloc here
    memcpy(result, s1, len1);
    memcpy(result+len1, s2, len2+1);//+1 to copy the null-terminator
    return result;
}

sha1 函数:

char *sha1( char *val ){

   int msg_length = strlen( val );
   int hash_length = gcry_md_get_algo_dlen( GCRY_MD_SHA1 );
   unsigned char hash[ hash_length ];
   char *out = (char *) malloc( sizeof(char) * ((hash_length*2)+1) );
   char *p = out;
   gcry_md_hash_buffer( GCRY_MD_SHA1, hash, val, msg_length );
   int i;
   for ( i = 0; i < hash_length; i++, p += 2 ) {
      snprintf ( p, 3, "%02x", hash[i] );
   }
   return out;
}

既然我问，它们是什么意思:

 //in real code you would check for errors in malloc here

提前致谢

Answer 1

sha1() 返回 char *，因此需要将 *m = sha1(as); 更改为 m = sha(as)；。 *m 的类型为 char，而不是 char *。