INT_MIN和编译器诊断 [英] INT_MIN and compiler diagnostic
问题描述
请参阅此测试程序：
1 #include< stdio.h>
2 #include< ; stdlib.h>
3 #include< limits.h>
4
5 int main（无效）
6 {
7 int y = -2147483648;
8 int x = INT_MIN;
9
10 printf（INT_MAX =％d INT_MIN =％d \ n，INT_MAX，
INT_MIN）;
11 printf（" x =％dy = ％d \ n"，x，y）;
12
13
14返回EXIT_SUCCESS;
15}
输出：
INT_MAX = 2147483647 INT_MIN = -2147483648
x = -2147483648 y = -2147483648
当我使用gcc编译它时，我在第7行得到诊断：
[pcg @ mylinux test] $ gcc -ansi -pedantic -Wall - o / tmp / x / tmp / xc
/ tmp / xc：函数`main''：
/ tmp / xc：7：警告：这个十进制常量是仅在ISO C90中无符号
然而，在我的系统中，INT_MIN确实是-2147483648，因为上面程序的输出建议为
。
是这种诊断有什么特别的原因[按照ANSI C]？
然而，第8行，逻辑上等同于第7行不会产生任何价值诊断。 INT_MIN在limits.h中定义为：
#define INT_MIN（-INT_MAX - 1）
#define INT_MAX 2147483647
谢谢，
pcg
gcc -ansi -pedantic -Wall -o / tmp / x / tmp / xc
/ tmp / xc：在函数main中：
/ tmp / xc：7：警告：此十进制常量仅为无符号在ISO C90
但是，在我的系统中，INT_MIN确实是-2147483648，因为上面程序的输出建议为
。
这个诊断是否有任何具体原因[按照ANSI C]？
但是，第8行，逻辑上等同于第7行不是
产生任何诊断。 INT_MIN在limits.h中定义为：
#define INT_MIN（-INT_MAX - 1）
#define INT_MAX 2147483647
谢谢，
pcg
blockquote>
2月28日，10：07，p_cricket _... @ yahoo.co .in写道：
当我使用gcc编译它时，我在第7行得到诊断：
[pcg @ mylinux test]
gcc -ansi -pedantic -Wall -o / tmp / x / tmp / xc
/ tmp / xc：函数`main'' ：
/ tmp / xc：7：警告：此十进制常数仅在ISO C90
中无符号
在我的（Sun sparc）上使用gcc与我得到完全相同的标志
xc：7：警告：十进制常量太大以至于无符号
程序输出是：
INT_MAX = 2147483647 INT_MIN = -2147483648
x = -2147483648 y = -2147483648
有趣的是它告诉我康斯坦t是未签名的，然后打印它
签名。
当你需要语言律师时，哪里是语言律师？！
Please see this test program:
1 #include <stdio.h>
2 #include <stdlib.h>
3 #include <limits.h>
4
5 int main (void)
6 {
7 int y = -2147483648;
8 int x = INT_MIN;
9
10 printf("INT_MAX = %d INT_MIN = %d\n", INT_MAX,
INT_MIN);
11 printf("x = %d y = %d\n", x, y);
12
13
14 return EXIT_SUCCESS;
15 }
Output:
INT_MAX = 2147483647 INT_MIN = -2147483648
x = -2147483648 y = -2147483648
When I compile this using gcc, I get a diagnostic on line 7:
[pcg@mylinux test]$gcc -ansi -pedantic -Wall -o /tmp/x /tmp/x.c
/tmp/x.c: In function `main'':
/tmp/x.c:7: warning: this decimal constant is unsigned only in ISO C90
However, on my system INT_MIN is indeed -2147483648 as
suggested by the output of above program.
Is there any specific reason for this diagnostic [as per ANSI C] ?
However, line 8, which is logically equivalent to line 7 does not
produce any diagnostic. INT_MIN is defined in limits.h as:
# define INT_MIN (-INT_MAX - 1)
# define INT_MAX 2147483647
Thanks,
pcg
解决方案gcc -ansi -pedantic -Wall -o /tmp/x /tmp/x.c
/tmp/x.c: In function `main'':
/tmp/x.c:7: warning: this decimal constant is unsigned only in ISO C90
However, on my system INT_MIN is indeed -2147483648 as
suggested by the output of above program.
Is there any specific reason for this diagnostic [as per ANSI C] ?
However, line 8, which is logically equivalent to line 7 does not
produce any diagnostic. INT_MIN is defined in limits.h as:
# define INT_MIN (-INT_MAX - 1)
# define INT_MAX 2147483647
Thanks,
pcg
On 28 Feb, 10:07, p_cricket_...@yahoo.co.in wrote:When I compile this using gcc, I get a diagnostic on line 7:
[pcg@mylinux test]
gcc -ansi -pedantic -Wall -o /tmp/x /tmp/x.c
/tmp/x.c: In function `main'':
/tmp/x.c:7: warning: this decimal constant is unsigned only in ISO C90On mine (Sun sparc) using gcc with exactly the same flags I get
x.c:7: warning: decimal constant is so large that it is unsigned
The program output is:
INT_MAX = 2147483647 INT_MIN = -2147483648
x = -2147483648 y = -2147483648
So amusingly, it tells me the constant is unsigned, then prints it
signed.
Where''s a language lawyer when you need one?!
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