在 C 中, < code> a 是字符常数, 被作为整数处理, 大小为 4, 在 C++ 中, 它被作为 < code>char 处理。 这是问题在此的重复 :

< a href=" "https://stackoverflow.com/ questions/2172943/ size-of-criacter-a-in-c-c" C/C/a> 字符大小(a)

In C a character literal (constant) has type int. So, consider the following program

#include <stdio.h>

main(int argc, char *argv[])
{
  printf("%zu
", sizeof( a ));
  printf("%zu
", sizeof( ab ));
  printf("%zu
", sizeof( abc ));
  printf("%zu
", sizeof( abcd ));

  printf("%u
",  a );
  printf("%u
",  ab );
  printf("%u
",  abc );
  printf("%u
",  abcd );

  printf("%x
",  a );
  printf("%x
",  ab );
  printf("%x
",  abc );
  printf("%x
",  abcd );

  printf("%c
",  a );
  printf("%c
",  ab );
  printf("%c
",  abc );
  printf("%c
",  abcd );
}

The first four statements all consider the literals as one character constant and they all print 4 == sizef(int), at least on gcc (Ubuntu 4.4.3-4ubuntu5.1) 4.4.3. Note that this compiler prints several warnings for the above program:

warning: multi-character character constant

Basically, a character literal specifies the four bytes making up an int, from left to right, higher-order byte first. The missing leading bytes are filled with 0. So, on my machine the second and third group of printf statements print

97
24930
6382179
1633837924
61
6162
616263
61626364

In the hexadecimal output you see the layout of the four characters in the literal (the ASCII codes from left to right): the a is mapped to the highest-order byte 0x61).

最后,第四组指纹:

a
b
c
d

i.e. the character literals are pushed on the stack as integers, but printf only prints the lowest byte of that int as a char.

C++ behaves in a similar way, but one-byte character literals are considered of type char, not int. The program

#include <iostream>

using namespace std;

main(int argc, char *argv[])
{
  cout << sizeof( a ) << endl;
  cout << sizeof( ab ) << endl;
  cout << sizeof( abc ) << endl;
  cout << sizeof( abcd ) << endl;

  cout <<  a  << endl;
  cout <<  ab  << endl;
  cout <<  abc  << endl;
  cout <<  abcd  << endl;
}

will compile using GCC and give a similar warning. Its output is different from that of C:

1
4
4
4
a
24930
6382179
1633837924

So one-byte character literals are treated as char, while multi-byte literals are treated as int.

<强度 > 说明

I ran my tests on a 32-bit Linux system on which an int has 4 bytes. It would be interesting to see what happens on other systems, e.g. on a 64-bit system.

<强 > EDIT

Fixed answer (thanks for the hint): character literals have type int in C, they are not cast to int.