轉換的char *爲char＆

Question

我有工作了下面的代碼罰款：

//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); 

，我想將其轉換爲避免額外的功能，做同樣的事情，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; 

但第二個不起作用。

* 如何將「char *」轉換爲「（char ）&」？

編輯：

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

您需要將數據複製到緩衝區。

//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.