Answer / susie

Answer :

A

Explanation:

Due to the assignment p[1] = ‘c’ the string becomes, “%c\n”.
Since this string becomes the format string for printf and
ASCII value of 65 is ‘A’, the same gets printed.

Answer / hunny kukreja

Answer:
65

Explanation:
As p is having "%d\n",so it has become format string
for printf,so same will get printed.i.e. number 65

In the following pgm add a stmt in the function fun such that the address of 'a' gets stored in 'j'. main(){ int * j; void fun(int **); fun(&j); } void fun(int **k) { int a =0; /* add a stmt here*/ }

1 Answers  

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main( ) { int a[ ] = {10,20,30,40,50},j,*p; for(j=0; j<5; j++) { printf(“%d” ,*a); a++; } p = a; for(j=0; j<5; j++) { printf(“%d ” ,*p); p++; } }

1 Answers  

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main(int argc, char **argv) { printf("enter the character"); getchar(); sum(argv[1],argv[2]); } sum(num1,num2) int num1,num2; { return num1+num2; }

1 Answers  

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Is the following statement a declaration/definition. Find what does it mean? int (*x)[10];

1 Answers  

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main() { char *cptr,c; void *vptr,v; c=10; v=0; cptr=&c; vptr=&v; printf("%c%v",c,v); }

1 Answers  

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main() { int i=5; printf(“%d”,i=++i ==6); }

1 Answers  

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void main() { int i=i++,j=j++,k=k++; printf(“%d%d%d”,i,j,k); }

1 Answers  

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Is there any difference between the two declarations, 1. int foo(int *arr[]) and 2. int foo(int *arr[2])

1 Answers  

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Write a routine to draw a circle (x ** 2 + y ** 2 = r ** 2) without making use of any floating point computations at all.

2 Answers   Mentor Graphics, Microsoft,

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void main() { printf(“sizeof (void *) = %d \n“, sizeof( void *)); printf(“sizeof (int *) = %d \n”, sizeof(int *)); printf(“sizeof (double *) = %d \n”, sizeof(double *)); printf(“sizeof(struct unknown *) = %d \n”, sizeof(struct unknown *)); }

1 Answers  

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main() { static char names[5][20]={"pascal","ada","cobol","fortran","perl"}; int i; char *t; t=names[3]; names[3]=names[4]; names[4]=t; for (i=0;i<=4;i++) printf("%s",names[i]); }

2 Answers  

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main( ) { static int a[ ] = {0,1,2,3,4}; int *p[ ] = {a,a+1,a+2,a+3,a+4}; int **ptr = p; ptr++; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); *ptr++; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); *++ptr; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); ++*ptr; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); }

2 Answers   Persistent,

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