char * 作为 C 中的引用

Question

如何传递参数如char *作为引用？

我的函数使用 malloc()

void set(char *buf)
{
    buf = malloc(4*sizeof(char));
    buf = "test";
}

char *str;
set(str);
puts(str);

Answer 1

你传递指针的地址:

void set(char **buf)
{
    *buf = malloc(5*sizeof(char));
    // 1. don't assign the other string, copy it to the pointer, to avoid memory leaks, using string literal etc.
    // 2. you need to allocate a byte for the null terminator as well
    strcpy(*buf, "test");
}

char *str;
set(&str);
puts(str);