Names of functions in two different files linked together must be unique
Answer: A
Correct answer: True
Once two files are linked together they share a single global namespace, so two external functions with the same name collide and the linker reports a duplicate symbol. The way to avoid it is to declare a function static, which confines it to its own file.
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What is the output of this C code?
double i;
int main()
{
printf("%g\n",i);
return 0;
}
Answer: A
Correct answer: 0.000000
A file-scope double is zero-initialised, so it holds 0.0. %g chooses the shorter of the fixed and exponential forms, and for zero the answer is 0 either way.
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What is the output of this C code?
void main()
{
m();
m();
}
void m()
{
static int x = 5;
x++;
printf("%d", x);
}
Answer: A
Correct answer: 6 7
A static local is initialised once, before the program starts, and keeps its value between calls - its storage lasts as long as the program even though its name is local. So the first call prints 6 and the second 7. An ordinary local would print 6 both times.
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What is the output of this C code?
int foo();
int main()
{
int i = foo();
}
foo()
{
printf("2 ");
return 2;
}
Answer: A
Correct answer: 2
The prototype says int foo(), and the definition, which gives no return type, defaults to int - so the two agree. The program compiles and prints 2.
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What is the output of this C code?
void main()
{
static int x = 3;
x++;
if (x <= 5)
{
printf("hi");
main();
}
}
Answer: D
Correct answer: hi hi
x is static, so it is initialised to 3 once and keeps its value across the recursive calls. It becomes 4 on the first pass (prints hi), 5 on the second (prints hi), and 6 on the third, where the test fails and the recursion stops. So hi appears twice. Had x been ordinary, it would have been reset to 3 on every call and the recursion would never have ended.
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Which of the following function declarations is illegal?
Answer: D
Correct answer: None of the mentioned
All three are legal. Declaring a function before main and defining it afterwards is the normal pattern; a stray semicolon after a definition is an empty declaration, which is harmless; and a declaration inside a function is allowed, its scope being that block.
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The keyword used to transfer control from a function back to the calling function is
Answer: D
Correct answer: return
return ends the function and hands control - and optionally a value - back to the caller. goto can only jump within the same function, and there is no go back keyword.
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What will be the output?
int main()
{
printf("%d", d++);
}
int d = 10;
Answer: D
Correct answer: Compile time error
d is declared after main, so at the point of use it is not in scope. Under C99 and later an undeclared identifier is an error; older compilers would have assumed int and then found the later definition consistent, but the modern answer is a compile error.
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The keyword used to transfer control from a function back to the calling function is
Answer: D
The keyword return is used to transfer control from a function back to the calling function.
Example:
#include
int add(int, int); /* Function prototype */
int main()
{
int a = 4, b = 3, c;
c = add(a, b);
printf("c = %d\n", c);
return 0;
}
int add(int a, int b)
{
/* returns the value and control back to main() function */
return (a+b);
}
Output: c = 7
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What is the output of this C code?
int x = 5;
void main()
{
int x = 3;
printf("%d", x);
{
int x = 4;
}
printf("%d", x);
}
Answer: A
Correct answer: 3 3
Here the inner int x = 4; is a declaration, so it creates a new variable that shadows the outer one for the length of the block and then disappears. main’s own x is untouched, and both printf calls show 3.
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