Question

使用运算符sizeof(对象或类)，显示各数据类型（char，int，short，long，float，double）所占的内存字节数

Answer 1

#include
<iostream.h>
　　int
main
(void)
　　{
　　cout
<<
"char
size
=
"
<<
sizeof(char)
<<
"
bytes\n";
　　cout
<<
"char*
size
=
"
<<
sizeof(char*)
<<
"
bytes\n";
　　cout
<<
"short
size
=
"
<<
sizeof(short)
<<
"
bytes\n";
　　cout
<<
"int
size
=
"
<<
sizeof(int)
<<
"
bytes\n";
　　cout
<<
"long
size
=
"
<<
sizeof(long)
<<
"
bytes\n";
　　cout
<<
"float
size
=
"
<<
sizeof(float)
<<
"
bytes\n";
　　cout
<<
"double
size
=
"
<<
sizeof(double)
<<
"
bytes\n";
　　cout
<<
"1.55
size
=
"
<<
sizeof(1.55)
<<
"
bytes\n";
　　cout
<<
"1.55l
size
=
"
<<
sizeof(1.55l)
<<
"
bytes\n";
　　cout
<<
"hello
size
=
"
<<
sizeof("hello")
<<
"
bytes\n";
　　}
　　我们运行这个程序，产生下面的结果（在作者的pc机上）：
　　char
size
=
1
bytes
　　char*
size
=
2
bytes
　　short
size
=
2
bytes
　　int
size
=
2
bytes
　　long
size
=
4
bytes
　　float
size
=
4
bytes
　　double
size
=
8
bytes
　　1.55
size
=
8
bytes
　　1.55l
size
=
10
bytes
　　hello
size
=
6
bytes

Answer 2

没错啊。。printf("\n%d\n",sizeof(int));可以打出改系统中int定义变量占多少个字节。一般的是4个