c - printf %f 如何处理 32 位 float

Question

%f printf 格式代码被指定为操作double 类型的值 [ source ].但是，一个简单的测试程序表明它也可以与 float 类型的值一起使用。这是如何工作的？

整数类型(例如 int 和 long long int)的等效情况“有效”，因为在小端机器上，32-位整数恰好与 64 位整数的低位字节重叠，因此只要高位为 0，您就会得到“正确”答案。

但这不可能是 float 和 double 的情况，因为浮点格式不能像这样互换。如果不进行(相当复杂的)转换为其他格式，您根本无法将浮点值打印为 double 值。尝试通过类型双关来做到这一点只会打印垃圾。

最重要的是，printf 是可变的。编译器不一定在编译时知道将使用什么格式说明符，只知道参数的类型。因此，我唯一可以推测的是，传递给可变参数函数的所有 float 值将无条件地升级为double。但令我难以置信的是，我可能已经用 C 语言编程了这么长时间，却不知道这一点。

C 如何在这里进行隐式强制转换？

来源:

#include <stdio.h>
#include <math.h>

int main() {
  float x[2] = {M_PI, 0.0};
  printf("value of x: %.16e\n", x[0]);
  printf("size of x: %lu\n", sizeof(x[0]));

  double *xp = (double *)&x[0];
  printf("value of *xp: %.16e\n", *xp);
  printf("size of *xp: %lu\n", sizeof(*xp));

  double y = M_PI;
  printf("value of y: %.16e\n", y);
  printf("size of y: %lu\n", sizeof(y));

  int i[2] = {1234, 0};
  printf("value of i: %lld\n", i[0]);
  printf("sizeof of i: %lu\n", sizeof(i[0]));

  long long *ip = (long long *)&i[0];
  printf("value of i: %lld\n", *ip);
  printf("sizeof of i: %lu\n", sizeof(*ip));

  return 0;
}

输出:

value of x: 3.1415927410125732e+00
size of x: 4
value of *xp: 5.3286462644388174e-315
size of *xp: 8
value of y: 3.1415926535897931e+00
size of y: 8
value of i: 1234
sizeof of i: 4
value of i: 1234
sizeof of i: 8

编译命令及版本:

$ gcc test_float.c -o test_float
$ gcc --version
gcc (Ubuntu 5.5.0-12ubuntu1~16.04) 5.5.0 20171010
Copyright (C) 2015 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

Answer 1

Therefore the only thing I can surmise is that all float values passed to a variadic function would be upgraded to double, unconditionally.

是的 - 完全正确。

来自C标准；

6.5.2.2.7 The ellipsis notation in a function prototype declarator causes argument type conversion to stop after the last declared parameter. The default argument promotions are performed on trailing arguments.

“默认参数提升”规则会将 float 提升为 double，相关部分为:

6.5.2.2.6 If the expression that denotes the called function has a type that does not include a prototype, the integer promotions are performed on each argument, and arguments that have type float are promoted to double.