main() { char *p="GOOD"; char a[ ]="

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Question

main()

      {

		char *p="GOOD";

      	char a[ ]="GOOD";

printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d",
sizeof(p), sizeof(*p), strlen(p));

      	printf("\n sizeof(a) = %d, strlen(a) = %d",
sizeof(a), strlen(a));

      }

Question Submitted By :: Susie

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Answer Posted By

Re: main() { char *p="GOOD"; char a[ ]="GOOD"; printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p), sizeof(*p), strlen(p)); printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a)); }

Answer
# 1

Answer : 

		sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4

      	sizeof(a) = 5, strlen(a) = 4

      Explanation:

		sizeof(p) => sizeof(char*) => 2

      	sizeof(*p) => sizeof(char) => 1

		Similarly,

      	sizeof(a) => size of the character array => 5

When sizeof operator is applied to an array it returns the
sizeof the array and it is not the same as the sizeof the
pointer variable. Here the sizeof(a) where a is the
character array and the size of the array is 5 because the
space necessary for the terminating NULL character should
also be taken into account.

Is This Answer Correct ?

0 Yes

0 No

Susie

Question	Asked @	Answers
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