在64位机器上，int 的大小应该是多少？

可能的复制品:
整数、长度等的大小
Int 的大小取决于编译器和/或处理器吗？
什么决定一个整数的大小

我用的是 64-bit机器。

$ uname -m
x86_64
$ file /usr/bin/file
/usr/bin/file: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.32, stripped
$

当我运行以下程序时，我得到了 sizeof(int)作为 4-bytes。

#include <stdio.h>


int main(void)
{
printf("sizeof(int) = %d bytes\n", (int) sizeof(int));


return 0;
}

如果我运行的是 16-、 32-和 64-位机，那么这是否意味着 integer的大小分别是 16-、 32-和 64-位？

在我的机器里，我发现 WORD_BIT是 32。它不应该是 64-bit机器上的 64吗？

$ getconf WORD_BIT
32
$

在上述情况下，sizeof(int)不应该是 64-bits(8 bytes)吗？

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小开

Not really. for backward compatibility it is 32 bits.
If you want 64 bits you have long, size_t or int64_t

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最佳答案

Doesn't have to be; "64-bit machine" can mean many things, but typically means that the CPU has registers that big. The sizeof a type is determined by the compiler, which doesn't have to have anything to do with the actual hardware (though it typically does); in fact, different compilers on the same machine can have different values for these.

小开

In C++, the size of int isn't specified explicitly. It just tells you that it must be at least the size of short int, which must be at least as large as signed char. The size of char in bits isn't specified explicitly either, although sizeof(char) is defined to be 1. If you want a 64 bit int, C++11 specifies long long to be at least 64 bits.

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Size of a pointer should be 8 byte on any 64-bit C/C++ compiler, but not necessarily size of int.


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