Answer / susie

Answer :

No

Explanation:

Functions can only pass pointers and not arrays. The numbers
that are allowed inside the [] is just for more readability.
So there is no difference between the two declarations.

#include<stdio.h> main() { int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} }; int *p,*q; p=&a[2][2][2]; *q=***a; printf("%d..%d",*p,*q); }

1 Answers  

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&#8206;#define good bad main() { int good=1; int bad=0; printf ("good is:%d",good); }

2 Answers  

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void main() { static int i; while(i<=10) (i>2)?i++:i--; printf(“%d”, i); }

2 Answers  

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how to test pierrot divisor

0 Answers  

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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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how to return a multiple value from a function?

5 Answers   Wipro,

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main(int argc, char *argv[]) { (main && argc) ? main(argc-1, NULL) : return 0; } a. Runtime error. b. Compile error. Illegal syntax c. Gets into Infinite loop d. None of the above

4 Answers   HCL, LG,

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main() { char *a = "Hello "; char *b = "World"; clrscr(); printf("%s", strcat(a,b)); } a. Hello b. Hello World c. HelloWorld d. None of the above

3 Answers   HCL,

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

1 Answers  

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main() { int i, j, *p; i = 25; j = 100; p = &i; // Address of i is assigned to pointer p printf("%f", i/(*p) ); // i is divided by pointer p } a. Runtime error. b. 1.00000 c. Compile error d. 0.00000

3 Answers   HCL,

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#define f(g,g2) g##g2 main() { int var12=100; printf("%d",f(var,12)); }

3 Answers  

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Under linux environment can u please provide a c code for computing sum of series 1-2+3-4+5......n terms and -1+2-3+4-5...n terms..

2 Answers  

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