关于运营商规模的疑问 [英] doubt regarding sizeof operator
问题描述
大家好，
我不能了解波纹管提到的行为
程序。
#include< stdio.h>
int main（void）
{
printf（" \ n Size =％d \ n"，sizeof 1< 0）;
返回0;
}
程序的输出是
尺寸= 0
但我的理解是sizeof返回类型的大小。在这个
的情况下，它不应该返回sizeof（int）吗？
但是如果我重写代码如下所述输出是4
#include< stdio.h>
int main（无效）
{
printf（" \ n Size =％d \ n"，sizeof（1< 0））;
返回0;
}
尺寸= 4
你能帮我解释一下这个问题吗？
问候
Somenath
somenath写道：
>
大家好，
我无法理解提到的波纹管的行为
程序。
#include< stdio.h>
int main（void）
{
printf（" \ n Size =％d \ n"，sizeof 1< 0）;
返回0;
}
该程序的输出是
尺寸= 0
但我的理解是sizeof返回类型的大小。在这个
的情况下，它是否应该返回sizeof（int）？
表达式sizeof 1< 0"是假的，因此是零。 sizeof binds
比''<''更紧，所以你写了（sizeof 1）< 0.
-
Ian Collins。
10月30日，4 44，somenath < somenath ... @ gmail.comwrote：
大家好，
我无法理解这种行为提到的轰鸣声
程序。
#include< stdio.h>
int main（无效）
{
printf（" \ n Size =％d \ n"，sizeof 1< 0）;
返回0;}
程序的输出是
大小= 0
但我的理解是sizeof返回大小在这个
的情况下，它不应该返回sizeof（int）吗？
但是如果我按照提到的那样重写代码，则输出为4
#include< stdio.h>
int main（无效）
{
printf（ " \ n Size =％d \ n"，sizeof（1< 0））;
返回0;
}
Siz e = 4
能帮助我理解这个问题吗？
问候
Somenath
printf（" \ n Size =％d \ n"，sizeof（1< 0））; // sizeof（bool）
printf（" \ n Size =％d \ n"，sizeof 1< 0）; //等于（（sizeof（int））< 0）
当然为零 - -false--
somenath< so ********* @ gmail.comwrites：
我无法理解提到的波纹管的行为
程序。
#include< stdio.h>
int main（无效）
{
printf（" \ n Size =％d \ n"，sizeof 1< 0）;
返回0;
}
程序的输出是
大小= 0
Ian Collins解释说，这正是应该是什么。
[...]
但如果我按照提到的那样重写代码，则输出为4
#include< stdio.h>
int main（void）
{
printf（" \ n Size =％d \ n"，sizeof（1< 0））;
返回0;
}
大小= 4
（1< 0）产生int类型的值（带值0），因此sizeof（1 <0）
产生sizeof（int）。显然你的系统上有4个。但是你
使用％d打印该值。格式，对于int类型是正确的，
但不适用于类型size_t。
对于像这样的小值，打印它的合理方法是
将其转换为int：
printf（" \ n Size =％d \ n"，（int）sizeof（1< 0） ）;
-
Keith Thompson（The_Other_Keith） ks *** @ mib.org < http://www.ghoti.net/~kst>
圣地亚哥超级计算机中心< *< http://users.sdsc.edu/ ~kst>
我们必须做点什么。这是事情。因此，我们必须这样做。
- Antony Jay和Jonathan Lynn，是部长
Hi All,
I am not able to understand the behavior of the bellow mentioned
program.
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof 1<0);
return 0;
}
The output of the program is
Size = 0
But my understanding is sizeof returns the size of the type .In this
case should it not returns sizeof (int) ?
But if I rewrite the code as mentioned bellow the output is 4
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof (1<0));
return 0;
}
Size = 4
Could you please help me to understand the problem?
Regards
Somenath
somenath wrote:>
Hi All,
I am not able to understand the behavior of the bellow mentioned
program.
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof 1<0);
return 0;
}
The output of the program is
Size = 0
But my understanding is sizeof returns the size of the type .In this
case should it not returns sizeof (int) ?
The expression "sizeof 1<0" is false, hence the zero. sizeof binds
tighter than ''<'', so you have written (sizeof 1) < 0.
--
Ian Collins.
On 10 30 , 4 44 , somenath <somenath...@gmail.comwrote:Hi All,
I am not able to understand the behavior of the bellow mentioned
program.
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof 1<0);
return 0;}
The output of the program is
Size = 0
But my understanding is sizeof returns the size of the type .In this
case should it not returns sizeof (int) ?
But if I rewrite the code as mentioned bellow the output is 4
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof (1<0));
return 0;
}
Size = 4
Could you please help me to understand the problem?
Regards
Somenathprintf("\n Size = %d \n",sizeof (1<0));//sizeof(bool)
printf("\n Size = %d \n",sizeof 1<0);// equal to ((sizeof (int))<0)
of course zero --false--
somenath <so*********@gmail.comwrites:I am not able to understand the behavior of the bellow mentioned
program.
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof 1<0);
return 0;
}
The output of the program is
Size = 0Which is exactly what it should be, as Ian Collins explained.
[...]
But if I rewrite the code as mentioned bellow the output is 4
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof (1<0));
return 0;
}
Size = 4(1<0) yields a value of type int (with the value 0), so sizeof (1<0)
yields sizeof(int). Apparently that''s 4 on your system. But you
print the value using a "%d" format, which is correct for type int,
but not for type size_t.
For a small value like this, a reasonable way to print it is to
convert it to int:
printf("\n Size = %d \n",(int)sizeof (1<0));
--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
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