Arjun Suresh (talk | contribs) (→{{Template:Author|Arjun Suresh|{{arjunweb}} }}) |
Arjun Suresh (talk | contribs) (→{{Template:Author|Arjun Suresh|{{arjunweb}} }}) |
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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 extended to 0002 and passed to printf. Now, in printf we used %d, and hence the output will be 2. | + | 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. | 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. | ||
<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 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.