Answer / susie

Answer :

Compiler error: Cannot modify a constant value.

Explanation:

p is a pointer to a "constant integer". But we
tried to change the value of the "constant integer".

Answer / mahe

5
pointer value does not change.so print thier value

Answer / jambalakadi pamba

here...p is a pointer which is pointing to a addresss which is constant....!!! so the output is 6

main() { int i=0; while(+(+i--)!=0) i-=i++; printf("%d",i); }

9 Answers   CSC, GoDB Tech, IBM,

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main() { char a[4]="HELL"; printf("%s",a); }

3 Answers   Wipro,

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int DIM(int array[]) { return sizeof(array)/sizeof(int ); } main() { int arr[10]; printf(“The dimension of the array is %d”, DIM(arr)); }

2 Answers   CSC,

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# include <stdio.h> int one_d[]={1,2,3}; main() { int *ptr; ptr=one_d; ptr+=3; printf("%d",*ptr); }

1 Answers  

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main() { signed int bit=512, mBit; { mBit = ~bit; bit = bit & ~bit ; printf("%d %d", bit, mBit); } } a. 0, 0 b. 0, 513 c. 512, 0 d. 0, -513

3 Answers   HCL, Logical Computers,

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What is the subtle error in the following code segment? void fun(int n, int arr[]) { int *p=0; int i=0; while(i++<n) p = &arr[i]; *p = 0; }

1 Answers  

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main() { int i=0; for(;i++;printf("%d",i)) ; printf("%d",i); }

1 Answers   Zoho,

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main() { extern int i; i=20; printf("%d",i); }

1 Answers   Value Labs,

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main() { int i; clrscr(); for(i=0;i<5;i++) { printf("%d\n", 1L << i); } } a. 5, 4, 3, 2, 1 b. 0, 1, 2, 3, 4 c. 0, 1, 2, 4, 8 d. 1, 2, 4, 8, 16

4 Answers   HCL,

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void main() { int c; c=printf("Hello world"); printf("\n%d",c); }

2 Answers  

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main() { int i=400,j=300; printf("%d..%d"); }

3 Answers  

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int i=10; main() { extern int i; { int i=20; { const volatile unsigned i=30; printf("%d",i); } printf("%d",i); } printf("%d",i); }

1 Answers  

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