c++ - scanf 中是否有保证的分配顺序？

Question

我遇到了一些代码，想知道它是否按预期工作只是侥幸，还是只是错误的做法。考虑以下 MCVE ( ideone ):

#include <cstdio>

struct dummyStruct
{
    unsigned short min[4];
    unsigned short max[4];
    int            dummyBuffer; // This just happens to be here as a real variable in the original code, not just as a buffer.
};


int main()
{
    dummyStruct db;
    // Note that the size of the short is assumed to be half of that of the %d specifier
    sscanf("  123,   456,  789,   112", "%d, %d, %d, %d", db.min+0, db.min+1, db.min+2, db.min+3);
    sscanf("29491, 29491, 29491, 29491", "%d, %d, %d, %d", db.max+0, db.max+1, db.max+2, db.max+3);
    db.dummyBuffer = 1234;
    printf("%hd, %hd, %hd, %hd\n", db.min[0], db.min[1], db.min[2], db.min[3]);
    printf("%hd, %hd, %hd, %hd\n", db.max[0], db.max[1], db.max[2], db.max[3]);
    printf("%d\n", db.dummyBuffer);

    return 0;
}

结构的内容是否由标准保证，或者这是未定义的行为？我在 N4810 中没有看到这一点。或者，如果我们颠倒变量的顺序，例如

printf("%hd, %hd, %hd, %hd\n", db.min[0], db.min[2], db.min[1], db.min[3]);

db.min 的内容是否有保证？参数的顺序(从左到右)是赋值的顺序吗？另请注意，即使已定义，我也不是在问为什么这是不好的做法。我也不需要告诉我不要使用 scanf 的注释。我不是。

Answer 1

您没有在 N4810 中看到任何提及，因为当涉及到 C 标准库时，规范主要被推迟到 “ISO/IEC 9899:2011，编程语言 — C”。如果我们看一下 N1570 (C11 草案)，它说的是关于 scanf函数族:

7.21.6.2 The fscanf function (emphasis mine)

10 Except in the case of a % specifier, the input item (or, in the case of a %n directive, the count of input characters) is converted to a type appropriate to the conversion specifier. If the input item is not a matching sequence, the execution of the directive fails: this condition is a matching failure. Unless assignment suppression was indicated by a *, the result of the conversion is placed in the object pointed to by the first argument following the format argument that has not already received a conversion result. If this object does not have an appropriate type, or if the result of the conversion cannot be represented in the object, the behavior is undefined.

所以你的样本工作确实是一个未定义行为的侥幸。