c - 字符串数组初始化

Question

这段代码交换给定数组的第一个和最后一个元素:

#include <stdio.h>

/* swap first and last element of array */
void fn(void *v, size_t length, size_t size)
{
    char *a = (char *)v;
    char *b = (char *)v + size * (length - 1);
    char temp;

    do {
        temp = *a;
        *a++ = *b;
        *b++ = temp;
    } while (--size > 0);
}

int main(void)
{
    int a[] = {0, 1, 2};
    double b[] = {0., 1., 2.};
    char c[][15] = {"Hello", ",", "this is a test"};

    fn(a, 3, sizeof(a[0]));
    fn(b, 3, sizeof(b[0]));
    fn(c, 3, sizeof(c[0]));
    printf("%d %d\n", a[0], a[2]);
    printf("%f %f\n", b[0], b[2]);
    printf("%s %s\n", c[0], c[2]);
    return 0;
}

输出:

2 0
2.000000 0.000000
this is a test Hello

我的问题是:

代码安全吗，保证c[0]被编译器初始化为

{'h', 'e', 'l', 'l', 'o', 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

或c[0]可能包含{'h', 'e', 'l', 'l', 'o', 0, [garbage]}?

编辑:如果c没有初始化呢？

int main(void)
{
    char c[3][15];

    strcpy(c[0], "Hello");
    strcpy(c[1], ",");
    strcpy(c[2], "this is a test");
    fn(c, 3, sizeof(c[0]));
    printf("%s %s\n", c[0], c[2]);
    return 0;
}

使用 fn 安全吗？ (c[0] 包含 9 个字节的垃圾)

Answer 1

来自 C11 标准第 6.7.9 章:

21 If there are fewer initializers in a brace-enclosed list than there are elements or members of an aggregate, or fewer characters in a string literal used to initialize an array of known size than there are elements in the array, the remainder of the aggregate shall be initialized implicitly the same as objects that have static storage duration

是的，c[0] 将被完全初始化。

这也是由 C99 标准(第 6.7.8/21 章)定义的。对于标准的早期版本，我不知道这一点(不再...... ;-))。