What is the output of this C code?
static int x = 5;
void main()
{
x = 9;
{
int x = 4;
}
printf("%d", x);
}
Answer: A
Correct answer: 9
x = 9; assigns to the file-scope static variable, and the inner block’s int x = 4; declares a different variable that vanishes at the closing brace. So the printf, back outside, shows 9.
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Which of the following function declarations is illegal?
Answer: D
Correct answer: All of the mentioned
All three are illegal for the same reason: a C identifier cannot begin with a digit. 1bhk and 2bhk are not valid names, whatever their parameter lists.
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Automatic variables are allocated space in the form of a:
Answer: A
Correct answer: stack
Automatic variables are allocated on the stack: space is reserved when the block is entered and released when it is left, which is what makes their lifetime match the block and what makes recursion possible - each call gets its own frame.
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What is the output of this C code?
int x = 5;
void main()
{
int x = 3;
m();
printf("%d", x);
}
void m()
{
x = 8;
n();
}
void n()
{
printf("%d", x);
}
Answer: A
Correct answer: 8 3
main declares a local x = 3, which shadows the file-scope x = 5 inside main only. m is outside that scope, so its x = 8 assigns to the global, and n prints that 8. Back in main, the local is still 3.
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The output of the code below is
int *m();
void main()
{
int *k = m();
printf("hello ");
printf("%d", k[0]);
}
int *m()
{
int a[2] = {5, 8};
return a;
}
Answer: C
Correct answer: hello followed by garbage value
a is a local array, and it ceases to exist when m returns - so the pointer returned is dangling. Reading k[0] reads memory that has been released and may already have been reused, giving whatever happens to be there. To return an array safely it must be static, or allocated with malloc, or supplied by the caller.
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What is the output of this C code?
int main()
{
void foo();
void f()
{
foo();
}
f();
}
void foo()
{
printf("2 ");
}
Answer: D
Correct answer: Depends on the compiler
The program tries to define one function inside another, which standard C does not allow - so strictly it should not compile. GCC, however, offers nested functions as an extension, and there it compiles and prints 2. The result therefore depends on which compiler is used.
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What will be the data type returned for the following function?
int func()
{
return (double)(char)5.0;
}
Answer: B
Correct answer: int
The declared return type wins. Whatever casting happens inside, the value is converted to the function’s return type on the way out - so func returns an int. Here 5.0 becomes a char, then a double, and finally an int at the return.
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Global variables are:
Answer: B
Correct answer: External
A variable declared at file scope has external linkage by default, so other files can reach it with an extern declaration. Adding static restricts it to the file in which it is defined.
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What is the output of this C code?
void main()
{
static int x;
printf("x is %d", x);
}
Answer: D
Correct answer: 0
A static variable, whether at file scope or inside a function, is zero-initialised if given no initialiser. So it prints 0 - unlike an ordinary automatic variable, which would hold junk.
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The output of the code below is
int *m()
{
int *p = 5;
return p;
}
void main()
{
int *k = m();
printf("%d", k);
}
Answer: A
Correct answer: 5
int *p = 5; puts the value 5 into a pointer - meaningless as an address, but the bit pattern is simply carried back and printed with %d as 5. The code compiles only with warnings, and would crash the moment anything dereferenced p.
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