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All questions of C Functions for Class 6 Exam
Determine output:main()
{
int i = abc(10);
printf("%d", --i);
}int abc(int i)
{
return(i++);
}- a)10
- b)9
- c)11
- d)None of these.
Correct answer is option 'B'. Can you explain this answer?
Determine output:
main()
{
int i = abc(10);
printf("%d", --i);
}
{
int i = abc(10);
printf("%d", --i);
}
int abc(int i)
{
return(i++);
}
{
return(i++);
}
a)
10
b)
9
c)
11
d)
None of these.
|
Mahesh Chavan answered |
The code is incomplete as there is a missing closing bracket after the "printf(". Please provide the complete code to determine the output.
Which of the following declaration is illegal?- a)char *str = “Best C programming classes by Sanfoundry”;
- b)char str[] = “Best C programming classes by Sanfoundry”;
- c)char str[20] = “Best C programming classes by Sanfoundry”;
- d)char[] str = “Best C programming classes by Sanfoundry”;
Correct answer is option 'D'. Can you explain this answer?
Which of the following declaration is illegal?
a)
char *str = “Best C programming classes by Sanfoundry”;
b)
char str[] = “Best C programming classes by Sanfoundry”;
c)
char str[20] = “Best C programming classes by Sanfoundry”;
d)
char[] str = “Best C programming classes by Sanfoundry”;
|
Sarita Singh answered |
char[] str is a declaration in Java, but not in C.
Which keyword is used to prevent any changes in the variable within a C program?- a)immutable
- b)mutable
- c)const
- d)volatile
Correct answer is option 'C'. Can you explain this answer?
Which keyword is used to prevent any changes in the variable within a C program?
a)
immutable
b)
mutable
c)
const
d)
volatile
|
|
Sarita Singh answered |
const is a keyword constant in C program.
What will be the output of the following C code?#include <stdio.h>
int main()
{
const int i = 10;
int *ptr = &i;
*ptr = 20;
printf("%d\n", i);
return 0;
}- a)Compile time error
- b)Compile time warning and printf displays 20
- c)Undefined behaviour
- d)10
Correct answer is option 'B'. Can you explain this answer?
What will be the output of the following C code?
#include <stdio.h>
int main()
{
const int i = 10;
int *ptr = &i;
*ptr = 20;
printf("%d\n", i);
return 0;
}
int main()
{
const int i = 10;
int *ptr = &i;
*ptr = 20;
printf("%d\n", i);
return 0;
}
a)
Compile time error
b)
Compile time warning and printf displays 20
c)
Undefined behaviour
d)
10
|
|
Sarita Singh answered |
Changing const variable through non-constant pointers invokes compiler warning.
Output:
$ cc pgm2.c
pgm2.c: In function ‘main’:
pgm2.c:5: warning: initialization discards qualifiers from pointer target type
$ a.out
20
Output:
$ cc pgm2.c
pgm2.c: In function ‘main’:
pgm2.c:5: warning: initialization discards qualifiers from pointer target type
$ a.out
20
Any C program- a)Must contain at least one function.
- b)Need not contain any function.
- c)Needs input data.
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Any C program
a)
Must contain at least one function.
b)
Need not contain any function.
c)
Needs input data.
d)
None of the above
|
|
Sarita Singh answered |
At least one function which must be included in any C program is main().
What will be printed when this program is executed?int f(int x)
{
if(x <= 4)
return x;
return f(--x);
}
void main()
{
printf("%d ", f(7));
}- a)4 5 6 7
- b)1 2 3 4
- c)4
- d)Syntax error
Correct answer is option 'C'. Can you explain this answer?
What will be printed when this program is executed?
int f(int x)
{
if(x <= 4)
return x;
return f(--x);
}
void main()
{
printf("%d ", f(7));
}
{
if(x <= 4)
return x;
return f(--x);
}
void main()
{
printf("%d ", f(7));
}
a)
4 5 6 7
b)
1 2 3 4
c)
4
d)
Syntax error
|
|
Sarita Singh answered |
In this recursive function call the function will return to main caller when the value of x is 4. Hence the output.
Which of the following format identifier can never be used for the variable var?#include <stdio.h>
int main()
{
char *var = "Advanced Training in C by Sanfoundry.com";
}- a)%f
- b)%d
- c)%c
- d)%s
Correct answer is option 'A'. Can you explain this answer?
Which of the following format identifier can never be used for the variable var?
#include <stdio.h>
int main()
{
char *var = "Advanced Training in C by Sanfoundry.com";
}
int main()
{
char *var = "Advanced Training in C by Sanfoundry.com";
}
a)
%f
b)
%d
c)
%c
d)
%s
|
|
Sarita Singh answered |
%c can be used to print the indexed position.
%d can still be used to display its ASCII value.
%s is recommended.
%f cannot be used for the variable var.
%d can still be used to display its ASCII value.
%s is recommended.
%f cannot be used for the variable var.
What will be the output of the following C code?#include <stdio.h>
void foo(const int *);
int main()
{
const int i = 10;
printf("%d ", i);
foo(&i);
printf("%d", i);
}
void foo(const int *i)
{
*i = 20;
}- a)Compile time error
- b)10 20
- c)Undefined value
- d)10
Correct answer is option 'A'. Can you explain this answer?
What will be the output of the following C code?
#include <stdio.h>
void foo(const int *);
int main()
{
const int i = 10;
printf("%d ", i);
foo(&i);
printf("%d", i);
}
void foo(const int *i)
{
*i = 20;
}
void foo(const int *);
int main()
{
const int i = 10;
printf("%d ", i);
foo(&i);
printf("%d", i);
}
void foo(const int *i)
{
*i = 20;
}
a)
Compile time error
b)
10 20
c)
Undefined value
d)
10
|
|
Sarita Singh answered |
Cannot change a const type value.
Output:
$ cc pgm1.c
pgm1.c: In function ‘foo’:
pgm1.c:13: error: assignment of read-only location ‘*i’
Output:
$ cc pgm1.c
pgm1.c: In function ‘foo’:
pgm1.c:13: error: assignment of read-only location ‘*i’
Will the following C code compile without any error?#include <stdio.h>
int main()
{
for (int k = 0; k < 10; k++);
return 0;
}- a)Yes
- b)No
- c)Depends on the C standard implemented by compilers
- d)Error
Correct answer is option 'C'. Can you explain this answer?
Will the following C code compile without any error?
#include <stdio.h>
int main()
{
for (int k = 0; k < 10; k++);
return 0;
}
int main()
{
for (int k = 0; k < 10; k++);
return 0;
}
a)
Yes
b)
No
c)
Depends on the C standard implemented by compilers
d)
Error
|
|
Sarita Singh answered |
Compilers implementing C90 do not allow this, but compilers implementing C99 allow it.
Output:
$ cc pgm4.c
pgm4.c: In function ‘main’:
pgm4.c:4: error: ‘for’ loop initial declarations are only allowed in C99 mode
pgm4.c:4: note: use option -std=c99 or -std=gnu99 to compile your code
Output:
$ cc pgm4.c
pgm4.c: In function ‘main’:
pgm4.c:4: error: ‘for’ loop initial declarations are only allowed in C99 mode
pgm4.c:4: note: use option -std=c99 or -std=gnu99 to compile your code
What is the output of this C code?int main()
{
void foo();
void f()
{
foo();
}
f();
}
void foo()
{
printf("2 ");
}- a) 2 2
- b) 2
- c) Compile time error
- d) Depends on the compiler
Correct answer is option 'D'. Can you explain this answer?
What is the output of this C code?
int main()
{
void foo();
void f()
{
foo();
}
f();
}
void foo()
{
printf("2 ");
}
{
void foo();
void f()
{
foo();
}
f();
}
void foo()
{
printf("2 ");
}
a)
2 2
b)
2
c)
Compile time error
d)
Depends on the compiler
|
|
Nishanth Iyer answered |
The output of this code will be an error because the function foo() is defined after it is called in the function f().
What will happen after compiling and running following code?main()
{
printf("%p", main);
}- a)Error
- b)Will make an infinite loop.
- c)Some address will be printed.
- d)None of these.
Correct answer is option 'C'. Can you explain this answer?
What will happen after compiling and running following code?
main()
{
printf("%p", main);
}
{
printf("%p", main);
}
a)
Error
b)
Will make an infinite loop.
c)
Some address will be printed.
d)
None of these.
|
|
Sarita Singh answered |
Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They will be printed as hexadecimal numbers.
main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They will be printed as hexadecimal numbers.
Which of the following is not a pointer declaration?- a)char a[10];
- b)char a[] = {‘1’, ‘2’, ‘3’, ‘4’};
- c)char *str;
- d)char a;
Correct answer is option 'D'. Can you explain this answer?
Which of the following is not a pointer declaration?
a)
char a[10];
b)
char a[] = {‘1’, ‘2’, ‘3’, ‘4’};
c)
char *str;
d)
char a;
|
|
Sarita Singh answered |
Array declarations are pointer declarations.
What will be the output of the following program code?main()
{
static int var = 5;
printf("%d ", var--);
if(var)
main();
}- a)5 5 5 5 5
- b)5 4 3 2 1
- c)Infinite Loop
- d)Compilation Error
Correct answer is option 'B'. Can you explain this answer?
What will be the output of the following program code?
main()
{
static int var = 5;
printf("%d ", var--);
if(var)
main();
}
{
static int var = 5;
printf("%d ", var--);
if(var)
main();
}
a)
5 5 5 5 5
b)
5 4 3 2 1
c)
Infinite Loop
d)
Compilation Error
|
|
Sarita Singh answered |
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.
Will the following C code compile without any error?#include <stdio.h>
int main()
{
int k;
{
int k;
for (k = 0; k < 10; k++);
}
}- a)Yes
- b)No
- c)Depends on the compiler
- d)Depends on the C standard implemented by compilers
Correct answer is option 'A'. Can you explain this answer?
Will the following C code compile without any error?
#include <stdio.h>
int main()
{
int k;
{
int k;
for (k = 0; k < 10; k++);
}
}
int main()
{
int k;
{
int k;
for (k = 0; k < 10; k++);
}
}
a)
Yes
b)
No
c)
Depends on the compiler
d)
Depends on the C standard implemented by compilers
|
|
Sarita Singh answered |
There can be blocks inside the block. But within a block, variables have only block scope.
Output:
$ cc pgm5.c
Output:
$ cc pgm5.c
What will be the output of the following C code?#include <stdio.h>
int main()
{
j = 10;
printf("%d\n", j++);
return 0;
}- a)10
- b)11
- c)Compile time error
- d)0
Correct answer is option 'C'. Can you explain this answer?
What will be the output of the following C code?
#include <stdio.h>
int main()
{
j = 10;
printf("%d\n", j++);
return 0;
}
int main()
{
j = 10;
printf("%d\n", j++);
return 0;
}
a)
10
b)
11
c)
Compile time error
d)
0
|
|
Shubham Gupta answered |
Explanation:
The code provided will result in a compile-time error. This is because the variable 'j' is being used without declaration.
Reasoning:
- In the code, the variable 'j' is used without being declared. This will result in a compile-time error as the compiler does not know the data type of 'j'.
Fix:
To fix this issue, you can declare 'j' as an integer before using it.
Example:
c
#include
int main()
{
int j = 10;
printf("%d\n", j++);
return 0;
}
By declaring 'j' as an integer, the code will no longer have a compile-time error and will output the value of 'j' (which is 10) before incrementing it due to the postfix increment operator (++).
What is the result of compiling and running this code?main()
{
char string[] = "Hello World";
display(string);
}void display(char *string)
{
printf("%s", string);
}- a)will print Hello World
- b)Compilation Error
- c)will print garbage value
- d)None of these.
Correct answer is option 'B'. Can you explain this answer?
What is the result of compiling and running this code?
main()
{
char string[] = "Hello World";
display(string);
}
{
char string[] = "Hello World";
display(string);
}
void display(char *string)
{
printf("%s", string);
}
{
printf("%s", string);
}
a)
will print Hello World
b)
Compilation Error
c)
will print garbage value
d)
None of these.
|
|
Sarita Singh answered |
Compiler Error : Type mismatch in redeclaration of function display
As the function display() is not defined before use, compiler will assume the return type of the function which is int(default return type). But when compiler will see the actual definition of display(), mismatch occurs since the function display() is declared as void. Hence the error.
As the function display() is not defined before use, compiler will assume the return type of the function which is int(default return type). But when compiler will see the actual definition of display(), mismatch occurs since the function display() is declared as void. Hence the error.
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