带条件（三元）表达式的运算符“ sizeof”

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问题描述

当给出三元表达式时，我很难理解 sizeof 的行为。

I have a hard time understanding sizeof's behaviour when given a ternary expression.

#define STRING "a string"

int main(int argc, char** argv)
{
  int a = sizeof(argc > 1 ? STRING : "");

  int b = sizeof(STRING);
  int c = sizeof("");

  printf("%d\n" "%d\n" "%d\n", a, b, c);

  return 0;
}

在此示例中（使用gcc 4.4.3和4.7.2测试，已编译对于 -std = c99 ），b为9（8个字符+隐式'\0'），c为1（隐式'\0'）。由于某种原因，a是 4 。

In this example (tested with gcc 4.4.3 and 4.7.2, compiled with -std=c99), b is 9 (8 characters + implicit '\0'), c is 1 (implicit '\0'). a, for some reason, is 4.

根据argc是否大于1，我希望a为9或1。我以为字符串文字可能会在传递给 sizeof 之前转换为指针，从而导致 sizeof（char *）是4。

I would expect a to be either 9 or 1, based on whether argc is greater than 1. I thought maybe the string literals get converted to pointers before being passed to sizeof, causing sizeof(char*) to be 4.

我尝试替换 STRING 和通过char数组...

I tried replacing STRING and "" by char arrays...

char x[] = "";
char y[] = "a string";
int a = sizeof(argc > 1 ? x : y);

...但是我得到了相同的结果（a = 4，b = 9，c = 1 ）。

... but I got the same results (a=4, b=9, c=1).

然后，我尝试进入，但是我没有发现任何明显的解释。出于好奇，我还尝试将x和y更改为其他类型：

Then I tried to dive into the C99 spec, but I did not find any obvious explanation in it. Out of curiosity I also tried changing changing x and y to other types:

char 和 long long int ：a变为8

两者都 short 或两者都 char ：a变为4

char and long long int: a becomes 8
both short or both char: a becomes 4

因此肯定存在某种转换，但我很难找到任何官方解释。我可以想象一下，算术类型会发生这种情况（我模糊地意识到当涉及到这些类型时会进行大量的提升），但是我不明白为什么三元表达式返回的字符串文字会转换为大小为4的东西。

So there's definitely some sort of conversion going on, but I struggle to find any official explanation. I can sort of imagine that this would happen with arithmetic types (I'm vaguely aware there's plenty of promotions going on when those are involved), but I don't see why a string literal returned by a ternary expression would be converted to something of size 4.

NB：在此计算机上 sizeof（int）== sizeof（foo *）== 4 。

NB: on this machine sizeof(int) == sizeof(foo*) == 4.

感谢指针专家。了解 sizeof 和？：的工作方式实际上使我尝试了更多类型的混搭，并观察了编译器的反应。我是为了完整起见而在编辑它们：

Thanks for the pointers guys. Understanding how sizeof and ?: work actually led me to try a few more type mashups and see how the compiler reacted. I'm editing them in for completeness' sake:

foo* x = NULL; /* or foo x[] = {} */
int  y = 0;    /* or any integer type */

int a = sizeof(argc > 1 ? x : y);

收益警告：条件表达式中的指针/整数类型不匹配[默认启用] 和 a == sizeof（foo *）。

带有 foo x []，bar y [] ， foo * x，bar * y 或 foo * x，y栏[] ，警告将变为指针类型不匹配。使用 void * 时没有警告。

With foo x[], bar y[], foo* x, bar* y or foo* x, bar y[], the warning becomes pointer type mismatch. No warning when using a void*.

float x = 0; /* or any floating-point type */
int   y = 0; /* or any integer type */

int a = sizeof(argc > 1 ? x : y);

没有警告，并且 a == sizeof（x）（即浮点类型）。

Yields no warning, and a == sizeof(x) (that is, the floating-point type).

float x = 0;    /* or any floating-point type */
foo*  y = NULL; /* or foo y[] = {} */

int a = sizeof(argc > 1 ? x : y);

收益率错误：条件表达式中的类型不匹配

如果我完全阅读过规范，将确保编辑此问题以指向相关部分。

If I ever read the spec completely I'll make sure to edit this question to point to the relevant parts.

表达式的运算符

带条件（三元）表达式的运算符“ sizeof”

问题描述

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