main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
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Answer / nithin
p has address of letter a. p++ means it will have address of
y. *p reffers to the character y. so pre-incrementing that
will result in letter z being printed.
| Is This Answer Correct ? | 19 Yes | 18 No |
Answer / kishor amballi
char *p = "ayqm";
this line points to a string which is in read only data segment.
the printf statement p++ will be pointing to y.
It dumps core when trying to assign z at that location because the memory pointed to string ayqm is READ ONLY.
| Is This Answer Correct ? | 4 Yes | 4 No |
Answer / govind verma
output will be error because p is a character pointer point to a constant string we cant modified it and in above program we try to modified at ++*(p++) ..so its lead to be error constatnt cant be modified......
| Is This Answer Correct ? | 0 Yes | 2 No |
Answer / mayank srivastava
printf("%c", p );-----> prints address of a, i.e. 1024 (say).
printf("%c", p++);-----> prints address of y, i.e. 1025 (say)
printf("%c", *(p++));-----> prints the char y,
printf("%c", ++*(p++));-----> prints the char z,
so final answer is z.
| Is This Answer Correct ? | 2 Yes | 5 No |
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3 Answers HCL, Logical Computers,
main() { int i=1; while (i<=5) { printf("%d",i); if (i>2) goto here; i++; } } fun() { here: printf("PP"); }
PROG. TO PRODUCE 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
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