C Programming and Technical

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

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C Other 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 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).