The scope of an automatic variable is:
Answer: D
Correct answer: Both (a) and (b)
An automatic variable is visible from its declaration to the end of the block that contains it, and that includes any blocks nested inside - unless one of them declares another variable of the same name, which shadows it.
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Functions can be called either by value or reference
Answer: A
Correct answer: True
C passes everything by value, but passing a pointer by value gives the function the caller’s address and so the effect of call by reference. Both styles are therefore available, which is why arrays - whose names decay to pointers - behave as though passed by reference.
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What is the output of the below code snippet?
#include <stdio.h>
main()
{
unsigned x = 5, y=&x, *p = y+0;
printf("%u",*p);
}
Answer: D
Correct answer: 5
y holds the address of x, and p is set to that same address, so *p is x itself - 5. (The code stores an address in an unsigned int, which is poor practice and provokes warnings; it happens to work only where a pointer and an unsigned int are the same width.)
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What is the output of this C code?
void m();
void n()
{
m();
}
void main()
{
void m()
{
printf("hi");
}
}
Answer: B
Correct answer: Compile time error
Two faults. Standard C does not allow a function to be defined inside another, and m is defined inside main here. And n, which calls m, is never called at all - so even if it compiled, nothing would be printed.
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Is an assignment statement that gives a value to a local static variable executed only once?
Answer: B
Correct answer: false
The initialisation of a static variable happens once, before the program starts. But an ordinary assignment to it inside the function is an executable statement like any other, and runs every time the function is called - which would defeat the whole point of making the variable static.
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Which of the following is the correct form of a function declaration?
Answer: A
Correct answer: return-type function-name(argument type);
A declaration (prototype) states the return type, the name and the parameter types, and ends with a semicolon - no body. Adding a body makes it a definition, which is a different thing.
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What is the output of this C code?
void foo();
int main()
{
void foo();
foo();
return 0;
}
void foo()
{
printf("2 ");
}
Answer: B
Correct answer: 2
The function is declared at file scope and again inside main. Repeating a declaration is harmless as long as the two agree, which they do, so the program compiles and prints 2.
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Which of the following storage classes supports the char data type?
Answer: D
Correct answer: All of the mentioned
char works with every storage class: auto char, static char and register char are all valid. Storage class and type are independent choices - the first says where and how long the object lives, the second what it holds.
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What is the output of this C code?
int main()
{
void foo(), f();
f();
}
void foo()
{
printf("2 ");
}
void f()
{
printf("1 ");
foo();
}
Answer: B
Correct answer: 1 2
Declaring void foo(), f(); inside main is legal - it declares them, without defining them, and the declaration is simply local in scope. f() is then called; it prints 1 and calls foo, which prints 2. Output: 1 2.
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What is the format identifier for “static a = 20.5;”?
Answer: B
Correct answer: %d
With no type given, static a = 20.5; falls back on the old implicit int, so the value is truncated to 20 and the correct specifier is %d. The declaration is legal only under the pre-C99 rules; a modern compiler will warn or reject it.
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