Answer / susie

Answer :

Compier error: Type mismatch in redeclaration of show.

Explanation:

When the compiler sees the function show it doesn't know
anything about it. So the default return type (ie, int) is
assumed. But when compiler sees the actual definition of
show mismatch occurs since it is declared as void. Hence the
error.

The solutions are as follows:

1. declare void show() in main() .

2. define show() before main().

3. declare extern void show() before the use of show().

Answer / ravneet kaur

ans: declaration terminated incorrectly.

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main() { main(); }

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int a=1; printf("%d %d %d",a++,a++,a); need o/p in 'c' and what explanation too

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main( ) { int a[ ] = {10,20,30,40,50},j,*p; for(j=0; j<5; j++) { printf(“%d” ,*a); a++; } p = a; for(j=0; j<5; j++) { printf(“%d ” ,*p); p++; } }

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main() { int i=400,j=300; printf("%d..%d"); }

3 Answers  

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main() { char *p="GOOD"; char a[ ]="GOOD"; printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p), sizeof(*p), strlen(p)); printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a)); }

1 Answers  

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int i=10; main() { extern int i; { int i=20; { const volatile unsigned i=30; printf("%d",i); } printf("%d",i); } printf("%d",i); }

1 Answers  

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