Main n i nnt c 2 nprintf c d c n nAnswer c 2 nExplanation Here unary minus or negation

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Previous Question: output? main() { char string[]="Hello World"; display(string); } void display(char *string) { printf("%s",string); } 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 Next Question: output?? #define int char main() {i nt i=65; printf("sizeof(i)=%d",sizeof(i)); } Answer: sizeof(i)=1 Explanation: Since the #define replaces the string int by the macro char 50. main() {i nt i=10; i=!i>14; Printf ("i=%d",i); } Answer: i=0 Explanation: In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

(#M40029230) C question C Keep an EYE

main()
{i
nt c=- -2;
printf("c=%d",c);
}
Answer: c=2;
Explanation: Here unary minus (or negation) operator is used twice. Same maths rules applies,
ie. minus * minus= plus.
Note: However you cannot give like --2. Because -- operator can only be applied to variables as
a decrement operator (eg., i--). 2 is a constant and not a variable: Asked In C MAN (11 years ago); Unsolved; Is this Puzzle helpful? (0) (0) Submit Your Solution

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Latest C Aptitude Question SOLUTION: main() {i nt c=- -2; printf("c=%d",c); } Answer: c=2; Explanation: Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus. Note: H

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(#M40029230) C question C Keep an EYE

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