为什么最小的INT，-2147483648，有型“长”？ [英] Why does the smallest int, −2147483648, have type 'long'?

问题描述

有关学校的项目，我已经到code中的C函数的printf。事情会pretty不错，但有一个问题，我无法找到一个很好的答案，所以我在这里。

For a school project, I've to code the C function printf. Things are going pretty well, but there is one question I can't find a good answer to, so here I am.

printf("PRINTF(d) \t: %d\n", -2147483648);

告诉我（ GCC -Werror -Wextra -Wall ）

   error: format specifies type 'int' but the argument has type 'long'
      [-Werror,-Wformat]
        printf("PRINTF(d) \t: %d\n", -2147483648);
                              ~~     ^~~~~~~~~~~
                              %ld

但是，如果我用一个int变量，一切都很顺利：

But if I use an int variable, everything is going well:

int i;

i = -2147483648;
printf("%d", i);

为什么？

我明白了很多分，是非常有趣的。总之，我想的printf 使用＆LT; STDARG.H＆GT; librairy等，在va_arg（va_list的AP，类型）也应该返回正确的类型。对于％d个和％I ，显然返回的类型是 INT 。它是否在改变什么？

I understood many points and were very interesting. Anyway, I guess printf is using the <stdarg.h> librairy and so, va_arg(va_list ap, type) should also return the right type. For %d and %i, obviously the type returned is an int. Does it change anything ?

推荐答案

在C， -2147483648 不是一个整型常量。 2147483648 是一个整型常量，而 - 只是一个一元运算符适用于它，产生一个恒定的前pression。值 2147483648 不适合在 INT （它的One过大， 2147483647 通常的最大整数），因此整型常量的类型长，这将导致你观察到的问题。如果你想提的下限值的 INT ，可以使用宏 INT_MIN 从 ＆LT;＆limits.h中GT; （便携式办法）或谨慎避免提及 2147483648 ：

In C, -2147483648 is not an integer constant. 2147483648 is an integer constant, and - is just a unary operator applied to it, yielding a constant expression. The value of 2147483648 does not fit in an int (it's one too large, 2147483647 is typically the largest integer) and thus the integer constant has type long, which causes the problem you observe. If you want to mention the lower limit for an int, either use the macro INT_MIN from <limits.h> (the portable approach) or carefully avoid mentioning 2147483648:

printf("PRINTF(d) \t: %d\n", -1 - 2147483647);

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