困惑于printf％f [英] Confused to printf %f
问题描述
昨天有人问我关于整数除法和printf
的问题，
我告诉他他应该做一个浮动/在你做任何事之前加倍
interger division。
但是他不这么认为，所以我试着做一些例子来解释，
但是，经过一些尝试，
当我尝试做一些整数常数除法时我很困惑，
我知道我应该做浮点除法类似3.0 / 6.0，
但我仍然感到困惑-0.124709来自以下
的例子？
我只是添加了一些虚拟变量，但结果将是显示一些
不同。
我溢出了什么吗？
对于不同的gcc来说情况有点不同/>
编译器，它真的很奇怪。
//示例1
#include< stdio.h>
int main（）
{
//双i;
//长a = 1，b = 4;
printf（"％f \ n"，3/7）;
printf（"％f\ n"，5/7）;
printf（"％。2f \\ \\ n"，5/7）;
printf（"％f\ n"，5.0 / 7.0）;
printf（"％d \ n" ;，4/7）;
返回0;
}
-bash-3.00 $ gcc zc
-bash-3.00 $ ./a.out
-0.124709
-0.124709
-0.12
0.714286
0
**********************
//示例2：
#include< stdio.h>
int main（）
{
加倍i;
长a = 1，b = 4;
printf（"％f\ n，3/7 ）;
printf（"％f \ nn"，5/7）;
printf（"％。2f \ nn"，5/7） ;
printf（"％f \ nn"，5.0 / 7.0）;
printf（"％d \ nn"，4/7）;
返回0;
}
-bash-3.00 $ gcc zc
-bash-3.00 $ ./a .out
0.000000
0.000000
0.00
0.71428 6
0
********
//示例3：
#include< stdio.h>
int main（）
{
double i;
/ / long a = 1，b = 4;
printf（"％f \ nn"，3/7）;
printf（"％f \ n，5/7）;
printf（"％。2f \ nn"，5/7）;
printf（"％f \ n" ;，5.0 / 7.0）;
printf（"％d \ nn"，4/7）;
返回0;
}
-bash-3.00 $ gcc -v
使用内置规格。
配置：FreeBSD / i386系统编译器
线程模型：posix
gcc版本3.4.2 [FreeBSD] 20040728
-bash-3.00 $ gcc zc
-bash-3.00 $ ./a.out
-0.124709
-0.124709
-0.12
0.714286
0
-bash-3.00 $ gcc41 zc
-bash-3.00 $ ./a.out
0.000000
0.000000
0.00
0.714286
0
谢谢任何信息！
gcc zc
-bash-3.00
./ a.out
-0.124709
-0.124709
-0.12
0.714286
0
**********************
//示例2：
#include< stdio.h>
int main（）
{
加倍i;
长a = 1，b = 4;
printf（"％f \ nn"，3/7）;
printf（"％f \ nn"，5/7）;
printf（"％。2f \ nn"，5/7）;
printf（"％f \ nn"，5.0 / 7.0）;
printf（"％d \ nn"，4/7）;
返回0;
}
-bash-3.00
gcc zc
-bash- 3.00
Someone asked me a question about integer division and printf
yesterday,
I tell him he should do a casting to float/double before you do any
interger division.
But he doesn''t think so, so I try to do some example to explain,
However, after some trying,
I confused when I try to do some integer constant division,
I know I should do float division something like 3.0/6.0,
but I''m still confused where the -0.124709 comes for the following
example?
I just add some dummy variables, but the result will show some
different.
did I overflow anything?
The situation is somehow a little different for a different gcc
compiler, it''s really wierd.
//example 1
#include <stdio.h>
int main()
{
//double i;
//long a=1,b=4;
printf("%f\n",3/7);
printf("%f\n",5/7);
printf("%.2f\n",5/7);
printf("%f\n",5.0/7.0);
printf("%d\n",4/7);
return 0;
}
-bash-3.00$ gcc z.c
-bash-3.00$ ./a.out
-0.124709
-0.124709
-0.12
0.714286
0
**********************
//example 2:
#include <stdio.h>
int main()
{
double i;
long a=1,b=4;
printf("%f\n",3/7);
printf("%f\n",5/7);
printf("%.2f\n",5/7);
printf("%f\n",5.0/7.0);
printf("%d\n",4/7);
return 0;
}
-bash-3.00$ gcc z.c
-bash-3.00$ ./a.out
0.000000
0.000000
0.00
0.714286
0
********
//example 3:
#include <stdio.h>
int main()
{
double i;
//long a=1,b=4;
printf("%f\n",3/7);
printf("%f\n",5/7);
printf("%.2f\n",5/7);
printf("%f\n",5.0/7.0);
printf("%d\n",4/7);
return 0;
}
-bash-3.00$ gcc -v
Using built-in specs.
Configured with: FreeBSD/i386 system compiler
Thread model: posix
gcc version 3.4.2 [FreeBSD] 20040728
-bash-3.00$ gcc z.c
-bash-3.00$ ./a.out
-0.124709
-0.124709
-0.12
0.714286
0
-bash-3.00$ gcc41 z.c
-bash-3.00$ ./a.out
0.000000
0.000000
0.00
0.714286
0
Thanks for any information!
gcc z.c
-bash-3.00
./a.out
-0.124709
-0.124709
-0.12
0.714286
0
**********************
//example 2:
#include <stdio.h>
int main()
{
double i;
long a=1,b=4;
printf("%f\n",3/7);
printf("%f\n",5/7);
printf("%.2f\n",5/7);
printf("%f\n",5.0/7.0);
printf("%d\n",4/7);
return 0;
}
-bash-3.00
gcc z.c
-bash-3.00
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