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Consider the following C code segment.
int a, b, c = 0;
void prtFun(void);
main()
{
static int a = 1; /* Line 1 */
prtFun();
a += 1;
prtFun();
printf("
%d %d ", a, b);
}
void prtFun(void)
{
static int a = 2; /* Line 2 */
int b = 1;
a += ++b;
printf("
%d %d ", a, b);
}
What output will be generated by the given code segment?
- 3 1
4 1
4 2
- 4 2
6 1
6 1
- 4 2
6 2
2 0
- 3 1
5 2
5 2
Correct answer: 4 2
6 2
2 0
Solution
In prtFun, static a starts 2: first call ++b=2, a=4 -> '4 2'; second call a=6 -> '6 2'. In main the local static a=1 becomes 2 after a+=1, and global b is still 0, printing '2 0'. Output is 4 2 / 6 2 / 2 0.
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