[[低级1(进程2710)与代码06]]奇怪的输出

Question

我试图测量一个C程序的输出，所以我运行调试器。

这是程序-：

#define swap(a, b) temp=a; a=b; b=temp;
#include <stdio.h>

main()
{
        int i, j, temp;
    i = 5;
    j = 10;
    temp = 0;
    if(i > j)
        swap(i, j);
    printf("%d %d %d",  i, j, temp);
}

该程序的输出为10 0 0。我不知道怎么可能。

我在GDB的第6、7、8、9、10行插入了断点。这就是我得到的-：

(gdb) run
Starting program: /home/pritishc/Documents/a.out 

Breakpoint 1, main () at ProbleminC.c:7

7   i = 5;

(gdb) print i

$1 = 0

(gdb) print j

$2 = 0

(gdb) continue

Continuing.

Breakpoint 3, main () at ProbleminC.c:8

8   j = 10;

(gdb) print i

$3 = 5

(gdb) print j

$4 = 0

(gdb) print temp

$5 = 32767

(gdb) continue

Continuing.

Breakpoint 4, main () at ProbleminC.c:9

9   temp = 0;

(gdb) print i

$6 = 5

(gdb) print j

$7 = 10

(gdb) print temp

$8 = 32767

(gdb) c

Continuing.

Breakpoint 5, main () at ProbleminC.c:10

10  if(i > j)

(gdb) print i

$9 = 5

(gdb) print j

$10 = 10

(gdb) print temp

$11 = 0

(gdb) c

Continuing.

10 0 0[Inferior 1 (process 2710) exited with code 06]

(gdb) print i

No symbol "i" in current context.

(gdb)

这到底是什么意思？为什么它会给我这样的输出？

Answer 1

定义宏的方法是：

#define swap(a, b) temp=a; a=b; b=temp;

预处理器将将您的if条件转换为：

if(i > j)
    temp=i;
i=j;
j=temp;

将宏更改为：

#define swap(a, b) { temp=a; a=b; b=temp; }

或者甚至可以使用do ... while循环：

#define swap(a, b) do { temp=a; a=b; b=temp; } while (0)

所观察到的6的退出代码是printf的返回值。引用man 3 printf

   Upon successful return, these functions return the number of characters
   printed (excluding the null byte used to end output to strings).

在您的main() (宁愿使它成为int main())中，添加

return 0;