Answer / susie

Answer :

Compiler error (in the line arr1 list = {0,1,2,3,4})

Explanation:

arr2 is declared of type array of size 5 of characters. So
it can be used to declare the variable name of the type
arr2. But it is not the case of arr1. Hence an error.

Rule of Thumb:

#defines are used for textual replacement whereas typedefs
are used for declaring new types.

main() { char *p="hai friends",*p1; p1=p; while(*p!='\0') ++*p++; printf("%s %s",p,p1); }

3 Answers  

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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() { 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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#include<stdio.h> main() { register i=5; char j[]= "hello"; printf("%s %d",j,i); }

2 Answers  

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How to use power function under linux environment.eg : for(i=1;i<=n;i++){ pow(-1,i-1)} since it alerts undefined reference to 'pow'.

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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main() { show(); } void show() { printf("I'm the greatest"); }

2 Answers  

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

1 Answers  

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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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Give a very good method to count the number of ones in a 32 bit number. (caution: looping through testing each bit is not a solution)

7 Answers   Microsoft,

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#include<stdio.h> main() { char s[]={'a','b','c','\n','c','\0'}; char *p,*str,*str1; p=&s[3]; str=p; str1=s; printf("%d",++*p + ++*str1-32); }

1 Answers  

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posted by surbhi just now main() { float a = 5.375; char *p; int i; p=(char*)&a; for(i=0;i<=3;i++) printf("%02x",(unsigned char) p[i]); } how is the output of this program is :: 0000ac40 please let me know y this output has come

2 Answers   GATE,

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