Determine Output:
void main()
{
char *p="hi friends", *p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s", p1);
}
void main()
{
char *p="hi friends", *p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s", p1);
}A. hi friends
B. ij!gsjfoet
C. hj grjeodt
D. None of These
Answer: Option B
Solution(By Examveda Team)
++*p++ will be parse in the given order :
1. *p that is value at the location currently pointed by p will be taken
2. ++*p the retrieved value will be incremented
3. when ; is encountered the location will be incremented that is p++ will be executed.
This Question Belongs to C Program >> C Miscellaneous
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Comments ( 4 )
Determine output:
void main()
{
int const *p=5;
printf("%d", ++(*p));
}
A. 6
B. 5
C. Garbage Value
D. Compiler Error
A. mmm nnn aaa
B. mmmm nnnn aaaa
C. Compiler Error
D. None of These
A. I hate Examveda
B. I love Examveda
C. Error
D. None of These
Determine Output:
void main()
{
static int var = 5;
printf("%d ", var--);
if(var)
main();
}
A. 5 5 5 5 5
B. 5 4 3 2 1
C. Infinite Loop
D. None of These
The question should be modified as
{
char p[20] = "hi friends", * p1;
p1 = p;
while (*p1 != '') ++*p1++;
printf("%sn", p);
}
Hello,
This code must provide a runtime error because p is pointed to a literal string, which is unmodifiable.
Thanks.
Give full ans in brief !please.
P1 pointer is not effected by the moment of p1(they are copy of the same string hi friends) hence the answer should be hi friends , isn't it ??