Answer / susie

Answer :

Compiler Error : Type mismatch in redeclaration of
function display

Explanation :

In third line, when the function display is
encountered, the compiler doesn't know anything about the
function display. It assumes the arguments and return types
to be integers, (which is the default type). When it sees
the actual function display, the arguments and type
contradicts with what it has assumed previously. Hence a
compile time error occurs.

Answer / pratibha

this problem of type mismatching can be solved by declaring display function as a first line inside main() so that compiler will know that display is of type void and then it will resolve type mismatch while encountering calling function of display.

How we will connect multiple client ? (without using fork,thread)

3 Answers   TelDNA,

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main() { int y; scanf("%d",&y); // input given is 2000 if( (y%4==0 && y%100 != 0) || y%100 == 0 ) printf("%d is a leap year"); else printf("%d is not a leap year"); }

1 Answers  

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main() { int i; printf("%d",scanf("%d",&i)); // value 10 is given as input here }

2 Answers   IBM,

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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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main() { int x=5; clrscr(); for(;x<= 0;x--) { printf("x=%d ", x--); } } a. 5, 3, 1 b. 5, 2, 1, c. 5, 3, 1, -1, 3 d. –3, -1, 1, 3, 5

2 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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what is the output of the below program & why ? #include<stdio.h> void main() { int a=10,b=20,c=30; printf("%d",scanf("%d%d%d",&a,&b,&c)); }

6 Answers   CSC, IIIT,

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Design an implement of the inputs functions for event mode

0 Answers   Wipro,

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what is oop?

3 Answers  

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Predict the Output: int main() { int *p=(int *)2000; scanf("%d",2000); printf("%d",*p); return 0; } if input is 20 ,what will be print

2 Answers  

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main() { int i=0; while(+(+i--)!=0) i-=i++; printf("%d",i); }

9 Answers   CSC, GoDB Tech, IBM,

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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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