我想使用子函数来复制char数组。它是这样的:
void NSV_String_Copy (char *Source, char *Destination)
{
int len = strlen(Source);
if (*Destination != NULL)
free(Destination);
Destination = malloc(len + 1);
memmove(*Destination, Source, len);
Destination[len] = '\0'; //null terminate
}
这样,我可以从main函数调用它并以这种方式执行操作:
char *MySource = "abcd";
char *MyDestination;
NSV_String_Copy (MySource, MyDestination);
但是,它不能按预期工作。请帮忙!
答案 0 :(得分:2)
C按值传递参数,这意味着您无法使用问题中的函数原型更改调用者的MyDestination
。以下是更新调用者MyDestination
副本的两种方法。
选项a)传递MyDestination
void NSV_String_Copy (char *Source, char **Destination)
{
int len = strlen(Source);
if (*Destination != NULL)
free(*Destination);
*Destination = malloc(len + 1);
memmove(*Destination, Source, len);
(*Destination)[len] = '\0'; //null terminate
}
int main( void )
{
char *MySource = "abcd";
char *MyDestination = NULL;
NSV_String_Copy(MySource, &MyDestination);
printf("%s\n", MyDestination);
}
选项b)从函数返回Destination
,并将其分配给MyDestination
char *NSV_String_Copy (char *Source, char *Destination)
{
if (Destination != NULL)
free(Destination);
int len = strlen(Source);
Destination = malloc(len + 1);
memmove(Destination, Source, len);
Destination[len] = '\0'; //null terminate
return Destination;
}
int main( void )
{
char *MySource = "abcd";
char *MyDestination = NULL;
MyDestination = NSV_String_Copy(MySource, MyDestination);
printf("%s\n", MyDestination);
}