打印返回结构的成员 [英] printing a member of a returned struct

问题描述

我无法打印一个从函数返回一个结构的成员：

I'm having trouble printing a member of a struct that is returned from a function:

#include <stdio.h>

struct hex_string
{
    char a[9];
};

struct hex_string to_hex_string_(unsigned x)
{
    static const char hex_digits[] = "0123456789ABCDEF";
    struct hex_string result;
    char * p = result.a;
    int i;
    for (i = 28; i >= 0; i -= 4)
    {
        *p++ = hex_digits[(x >> i) & 15];
    }
    *p = 0;
    printf("%s\n", result.a);   /* works */
    return result;
}

void test_hex(void)
{
    printf("%s\n", to_hex_string_(12345).a);   /* crashes */
}

在的的printf 通话to_hex_string _ 打印正确的结果，但的printf 在 test_hex 崩溃我的程序。到底为什么会这样？它是一个终身的问题，或者是其他什么东西？

The printf call inside to_hex_string_ prints the correct result, but the printf call inside test_hex crashes my program. Why exactly is that? Is it a lifetime issue, or is it something else?

当我更换的printf 与看跌期权（to_hex_string_（12345）。A），我得到一个编译器调用错误：

When I replace the printf call with puts(to_hex_string_(12345).a), I get a compiler error:

invalid use of non-lvalue array

这是怎么回事？

推荐答案

有是在C规则，很少生效，其中规定：

There is a rule in C which seldom comes into effect, which states:

如果试图修改的函数调用或到结果
  下一个序列点之后访问它的行为是不确定的。 <子>（C99§6.5.2.2）

If an attempt is made to modify the result of a function call or to access it after the next sequence point, the behavior is undefined. _{(C99 §6.5.2.2)}

在这种情况下，有参数的printf（）进行评估，并在的printf（）函数本身执行。您传递给的printf指针（）是一个指针的返回值本身的的元素 - 当的printf（ ）试图通过该指针访问字符串，你会得到你的崩溃。

In this case, there is a sequence point after the arguments to printf() are evaluated and before the printf() function itself executes. The pointer you pass to printf() is a pointer to an element of the return value itself - and when printf() tries to access the string through that pointer, you get your crash.

这问题是很难碰到的，因为一个函数值不是左值，所以你不能直接拿一个指向它＆放大器;

This issue is hard to run into, because a function value isn't an lvalue so you can't directly take a pointer to it with &.

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