Answer / susie

Answer :

Yes

Explanation:

The typename aType is known at the point of declaring the
structure, because it is already typedefined.

main() { int *j; { int i=10; j=&i; } printf("%d",*j); }

9 Answers   HCL, Wipro,

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int i; main(){ int t; for ( t=4;scanf("%d",&i)-t;printf("%d\n",i)) printf("%d--",t--); } // If the inputs are 0,1,2,3 find the o/p

2 Answers  

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How can you relate the function with the structure? Explain with an appropriate example.

0 Answers  

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#include<stdio.h> void fun(int); int main() { int a; a=3; fun(a); printf("\n"); return 0; } void fun(int i) { if(n>0) { fun(--n); printf("%d",n); fun(--n); } } the answer is 0 1 2 0..someone explain how the code is executed..?

1 Answers   Wipro,

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main() { int i=10,j=20; j = i, j?(i,j)?i:j:j; printf("%d %d",i,j); }

2 Answers   Adobe, CSC,

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#define SQR(x) x * x main() { printf("%d", 225/SQR(15)); } a. 1 b. 225 c. 15 d. none of the above

3 Answers   HCL,

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main() { int i=-1,j=-1,k=0,l=2,m; m=i++&&j++&&k++||l++; printf("%d %d %d %d %d",i,j,k,l,m); }

1 Answers  

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#include<stdio.h> #include<conio.h> void main() { int a=(1,2,3,(1,2,3,4); switch(a) { printf("ans:"); case 1: printf("1");break; case 2: printf("2");break; case 3: printf("1");break; case 4: printf("4");break; printf("end"); } getch(); }

0 Answers  

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void main() { int x,y=2,z; z=(z*=2)+(x=y=z); printf("%d",z); }

4 Answers  

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void main() { int i=5; printf("%d",i+++++i); }

3 Answers  

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How to access command-line arguments?

4 Answers  

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What is wrong with the following code? int *foo() { int *s = malloc(sizeof(int)100); assert(s != NULL); return s; }

1 Answers  

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