Answer / susie

Answer :

contents of zzz.c followed by an infinite loop

Explanation:

The condition is checked against EOF, it should be checked
against NULL.

#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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how to concatenate the two strings

1 Answers  

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

4 Answers   Corporate Society, HCL,

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main() { int x=5; clrscr(); for(;x==0;x--) { printf("x=%d\n”", x--); } } a. 4, 3, 2, 1, 0 b. 1, 2, 3, 4, 5 c. 0, 1, 2, 3, 4 d. none of the above

3 Answers   HCL,

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Write a program to implement the motion of a bouncing ball using a downward gravitational force and a ground-plane friction force. Initially the ball is to be projected in to space with a given velocity vector

2 Answers  

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Which one is taking more time and why ? :/home/amaresh/Testing# cat time.c //#include <stdio.h> #define EOF -1 int main() { register int c; while ((c = getchar()) != EOF) { putchar(c); } return 0; } ------------------- WIth stdio.h:- :/home/amaresh/Testing# time ./time_header hi hi hru? hru? real 0 m4.202s user 0 m0.000s sys 0 m0.004s ------------------ Witout stdio.h and with #define EOF -1 =================== /home/amaresh/Testing# time ./time_EOF hi hi hru? hru? real 0 m4.805s user 0 m0.004s sys 0 m0.004s -- From above two case , why 2nd case is taking more time ?

0 Answers  

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main() { char *p = “ayqm”; printf(“%c”,++*(p++)); }

29 Answers   IBM, TCS, UGC NET, Wipro,

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#define square(x) x*x main() { int i; i = 64/square(4); printf("%d",i); }

4 Answers   Google, HCL, Quick Heal, WTF,

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1. const char *a; 2. char* const a; 3. char const *a; -Differentiate the above declarations.

3 Answers  

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Derive expression for converting RGB color parameters to HSV values

1 Answers  

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main() { char i=0; for(;i>=0;i++) ; printf("%d\n",i); }

2 Answers  

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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); }

2 Answers   CNSI,

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