转换的char *为char＆

Question

我有工作了下面的代码罚款：

//function1 for decoding base64 
int base64_decode (const char *base64, char *to) { /*function*/ } 

//code which is working 
buf_struct tmpbuf;//structure 
base64_decode(buffer, (char *)&tmpbuf); 

，我想将其转换为避免额外的功能，做同样的事情，到：

//function2 for decoding base64 
char *unbase64(unsigned char *input, int length) { /*function*/ } 

//code needs to be modified 
buf_struct tmpbuf;//structure 
char *unbase = unbase64(buffer, strlen(buffer)); 
unbase = (char *)&tmpbuf; 

但第二个不起作用。

* 如何将“char *”转换为“（char ）&”？

编辑：

char *unbase; 
unbase = malloc(strlen(buffer) + 1); 
memset(unbase, 0, strlen(buffer) + 1); 
//unbase = unbase64(buffer, strlen(buffer)); 
base64_decode(buffer, unbase); 
fprintf(stderr,"unbase: %s\n",unbase); 
strcpy((char *)&tmpbuf, unbase); 

Answer 1

您需要将数据复制到缓冲区。

//code needs to be modified 
buf_struct tmpbuf;//structure 
char *unbase = unbase64(buffer, strlen(buffer)); 
strcpy((char *)&tmpbuf, unbase); 

// Depending on the contract for unbase64 you may need to free() unbase here.