What linkage does automatic variables have?
Answer: C
Correct answer: No linkage
An automatic variable has no linkage: its name refers to a fresh object on each entry to the block and cannot be referred to from anywhere else - not from another file, not even from another function. Only file-scope names have linkage, internal or external.
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What is the output of this C code?
void main()
{
int x = 3;
{
x = 4;
printf("%d", x);
}
}
Answer: A
Correct answer: 4
The inner block contains an assignment, not a declaration, so it changes the x already in scope. The value printed is 4.
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What is the return type of the function sqrt()?
Answer: C
Correct answer: double
sqrt is declared in <math.h> as double sqrt(double) - it takes and returns a double. C99 added sqrtf and sqrtl for float and long double, but the return type of sqrt itself does not vary with the argument.
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What is the output of this C code?
void foo();
int main()
{
void foo(int);
foo(1);
return 0;
}
void foo(int i)
{
printf("2 ");
}
Answer: A
Correct answer: 2
The declaration inside main - void foo(int) - is the one in scope at the call, so foo(1) matches it, and it agrees with the definition. The outer void foo() declares a function with an unspecified parameter list rather than none, so it does not conflict. The program prints 2.
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What is the output of code given below?
enum m{JAN, FEB, MAR};
enum m foo();
int main()
{
enum m i = foo();
printf("%d\n", i);
}
int foo()
{
return JAN;
}
Answer: A
Correct answer: Compile time error
The prototype declares enum m foo(); but the definition is written int foo(). The two conflict, and the compiler rejects the program - even though an enumeration is represented as an integer, the declared types must match.
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Which variable has the longest scope?
int b;
int main()
{
int c;
return 0;
}
int a;
Answer: B
Correct answer: b
Scope runs from the point of declaration onwards, so b, declared at the very top, is visible through the whole file - the longest scope of the three. a is declared at the bottom and so is visible almost nowhere, and c is local to main.
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What is the scope of an external variable?
Answer: D
Correct answer: From the point of declaration to the end of the file being compiled
Scope runs from the point of declaration to the end of the file - a variable declared at the bottom of a file is not visible to functions above it. Being reachable from other files is a matter of linkage, which is a different property; an extern declaration in the other file is what brings it into scope there.
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What is the scope of a function?
Answer: D
Correct answer: From the point of declaration to the end of the file being compiled
A function’s name is in scope from its declaration to the end of the file - which is exactly why prototypes are put at the top or in a header. Its linkage is external, so other files can call it once they have declared it.
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What is the output of this C code?
int *i;
int main()
{
if (i == NULL)
printf("true\n");
return 0;
}
Answer: A
Correct answer: true
i is at file scope with no initialiser, so it is zero-initialised - and for a pointer that means the null pointer. The comparison with NULL therefore succeeds and true is printed.
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What is the output of the following program?
#include <stdio.h>
void swap(int m, int n)
{
int x = m;
m = n;
n = x;
}
main()
{
int x=5, y=3;
swap(x,y);
printf("%d %d", x, y);
}
Answer: B
Correct answer: 5 3
C passes arguments by value, so swap receives copies and exchanges the copies. The originals in main are untouched, and the output is 5 3. To swap the caller’s variables the function must take pointers and be called as swap(&x, &y).
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