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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Is initialization mandatory for local static variables?
Answer: B
Correct answer: NO
Initialisation is optional. A static variable with no initialiser is zero-initialised automatically, since it has static storage duration - unlike an automatic variable, which would hold junk. (If an initialiser is given, it must be a constant expression.)
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What are the uses of functions?
Answer: E
Correct answer: All of the above
Functions do all four. They let a block of code be written once and used many times; they make a program clearer by giving each task a name; they can be collected into libraries and reused across programs; and because each can be tested on its own, they make debugging far easier.
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What is the output of this C code?
int x = 5;
void main()
{
int x = 3;
printf("%d", x);
{
x = 4;
}
printf("%d", x);
}
Answer: D
Correct answer: 3 4
The inner x = 4; is an assignment, not a declaration, so it changes the x already in scope - main's local. The first printf shows 3 and the second, after the block, shows 4. The file-scope x = 5 is shadowed throughout and never seen.
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Functions cannot return a floating point number
Answer: B
Correct answer: No
A function may return any type it likes, floating point included - double sqrt(double) is the obvious example. The statement is false.
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What is the output of this C code?
double foo();
int main()
{
foo();
return 0;
}
foo()
{
printf("2 ");
return 2;
}
Answer: B
Correct answer: Compile time error
The prototype declares double foo(), but the definition omits the return type and so defaults to int. The two conflict, and the compiler rejects the program.
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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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