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So, the first two elements of the array will be stored in memory as follows: | So, the first two elements of the array will be stored in memory as follows: | ||
| − | 1000: 0 0 0 0 0 0 0 | + | 1000: 0 2 0 0 0 0 0 0 (hexa decimal representation and assuming int takes 4 bytes and assuming least significant byte is stored at lowest address(little-endian)) |
| − | 1004: 0 0 0 0 0 0 0 | + | 1004: 0 3 0 0 0 0 0 0 |
| − | Now, %d, *p | + | Now, %d, *p. |
| + | Here, *p is a character which is just a byte. So, *p value here is the byte at address 1000 which is 02 (in hex) and this is passed to printf. In C language, when a char is passed to a function actually it's int value as given by the char code (usually ASCII) is passed. So, here 02 is sign-extended to 0002 and passed to printf. Now, in printf we used %d, and hence the output will be 2. | ||
| + | |||
| + | When p is incremented, it will become to 1001, as p is a char pointer (for pointer addition, sizeof datatype pointed to is added and sizeof char is 1). So, now the value of *p is 00 and as explained above printf will print 0 for this. | ||
1001: 0 0 0 0 0 0 0 0 | 1001: 0 0 0 0 0 0 0 0 | ||
So, output will be 0. | So, output will be 0. | ||
| − | + | On a big-endian machine output will be 0 0. So, correct answer should be architecture dependent. | |
{{Template:FBD}} | {{Template:FBD}} | ||
<syntaxhighlight lang="c" name="pointer_1"> int main() {
GATECSE int arr[3] = {2, 3, 4};
char *p;
p = (char*)arr;
printf("%d ", *p);
p = p+1;
printf("%d\n", *p);
return 0;
}
</syntaxhighlight>
(A) 2 3
(B) 2 0
(C) 1 0
(D) Garbage value
Assume start location of arr is 1000. So, the first two elements of the array will be stored in memory as follows:
1000: 0 2 0 0 0 0 0 0 (hexa decimal representation and assuming int takes 4 bytes and assuming least significant byte is stored at lowest address(little-endian)) 1004: 0 3 0 0 0 0 0 0
Now, %d, *p. Here, *p is a character which is just a byte. So, *p value here is the byte at address 1000 which is 02 (in hex) and this is passed to printf. In C language, when a char is passed to a function actually it's int value as given by the char code (usually ASCII) is passed. So, here 02 is sign-extended to 0002 and passed to printf. Now, in printf we used %d, and hence the output will be 2.
When p is incremented, it will become to 1001, as p is a char pointer (for pointer addition, sizeof datatype pointed to is added and sizeof char is 1). So, now the value of *p is 00 and as explained above printf will print 0 for this. 1001: 0 0 0 0 0 0 0 0
So, output will be 0.
On a big-endian machine output will be 0 0. So, correct answer should be architecture dependent.
<syntaxhighlight lang="c" name="pointer_1"> int main() {
int arr[3] = {2, 3, 4};
char *p;
p = (char*)arr;
printf("%d ", *p);
p = p+1;
printf("%d\n", *p);
return 0;
}
</syntaxhighlight>
(A) 2 3
(B) 2 0
(C) 1 0
(D) Garbage value
Assume start location of arr is 1000. So, the first two elements of the array will be stored in memory as follows:
1000: 0 0 0 0 0 0 0 2 (hexa decimal representation and assuming int takes 4 bytes and assuming least significant byte is stored at highest address(little-endian)) 1004: 0 0 0 0 0 0 0 3
Now, %d, *p will print the first 4 bytes starting at 1000, that is 2. When p is incremented, it will go to 1001, as p is a char pointer.
1001: 0 0 0 0 0 0 0 0
So, output will be 0.