GATE CS 2015 Set 1 — Question 62
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Programming & Data Structures → Arrays & Strings → Matrix Operations
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Question
What is the output of the following C code? Assume that the address of x is 2000 (in decimal) and an integer requires four bytes of memory.
int main () {
unsigned int x[4][3] =
{{1,2,3},{4,5,6},{7,8,9},{10,11,12}};
printf("%u, %u, %u", x+3, *(x+3), *(x+2)+3);
}
Correct answer
(A) 2036, 2036, 2036
Solution
The array
x is defined as unsigned int x[4][3]. The base address is 2000, and sizeof(int) = 4.1.
x+3: x is a pointer to an array of 3 integers (type int (*)[3]). Incrementing it by 3 moves the pointer by .- Size of one row = bytes.
- Address = .
*(x+3): This dereferences the pointer to the 3rd row, giving x[3]. In C, an array name used in an expression decays to a pointer to its first element. So x[3] decays to a pointer to x[3][0] (type int *).- The address is the same as the start of the 3rd row: 2036.
*(x+2)+3: *(x+2) is x[2], which points to the start of the 2nd row (index 2). Address = . Adding 3 to this int * pointer moves it by bytes.- Address = .
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