Can You Pass This C Language Coding Test?
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Void main() { int const * p=5; printf("%d",++(*p)); }
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COMPILER ERROR
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6
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ADDRESS VALUE
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5
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NONE
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C is a programming language which was developed by Dennis Ritchie. C language incorporates low-level access to memory, clean style and simple set of keywords which makes 'C' language a perfect system programming like Operating system. This quiz has been designed to test your programming skills and basic knowledge about C language. There is a set of 10 questions. See moreRead the questions carefully and answer. So, let's try out the quiz. All the best!
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- 2.
Main() { char s[ ]="man"; int i; for(i=0;s[ i ];i++) printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]); }
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Mmm aaaa nnnnn
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Mmmmm aaa nnnn
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Mmmm aaaa nnnn
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Mmm aaa nnn
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NONE
Correct Answer
A. Mmmm aaaa nnnnExplanation
The given code is a C program that prints a pattern based on the characters in the string "man". The for loop iterates through each character in the string. Inside the loop, the printf statement is used to print the characters in different ways.
In the first iteration of the loop, the characters 'm', 'a', and 'n' are printed using different methods: s[i], *(s+i), *(i+s), and i[s].
The output of the program will be "mmmm", "aaaa", and "nnnn" printed on separate lines. Therefore, the correct answer is "mmmm aaaa nnnn".Rate this question:
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- 3.
6. main() { extern int i; i=20; printf("%d",i); }
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LINKER ERROR
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COMPILER ERROR
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20
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ADDRESS
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NONE
Correct Answer
A. LINKER ERRORExplanation
The code is trying to access a variable "i" that is declared as extern, which means it is defined in another file. However, there is no definition or declaration of "i" in the current file. This leads to a linker error because the linker is unable to find the definition of "i" and link it with the code. Therefore, the correct answer is LINKER ERROR.Rate this question:
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- 4.
Predict the output of the following C snippets int main() { int main = 56; printf("%d", main); return 0; }
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Compiler Error
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Depends on the compiler
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56
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None of above
Correct Answer
A. 56Explanation
main is not a keyword in C.Rate this question:
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- 5.
What is the output of this C code? #include int main() { float a = 3.14, *fptr; fptr = &a; printf("Size of Float Pointer : %d", sizeof(fptr)); return (0); }
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Size of Float Pointer : 2
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Size of Float Pointer : 4
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Size of Float Pointer : 6
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Size of Float Pointer : 8
Correct Answer
A. Size of Float Pointer : 8Explanation
The correct answer is "Size of Float Pointer : 8". In this code, a float variable "a" is declared and assigned a value of 3.14. Then, a float pointer "fptr" is declared and assigned the address of "a". The sizeof() function is used to determine the size of the float pointer "fptr", which is 8 bytes on most systems.Rate this question:
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- 6.
What is the output of the below c code? #include<stdio.h> int main() { char ch; if(ch = printf("")) printf("It matters\n"); else printf("It doesn't matters\n"); return 0; }
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It matters
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It doesn’t matters
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Run time error
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Nothing
Correct Answer
A. It doesn’t mattersExplanation
printf returns the number of characters successfully written on output.Hence ch=0 and else statement executes.Rate this question:
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- 7.
Which one do you like? int main() { fork(); fork(); printf("Hello world\n"); }
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1
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2
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4
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8
Correct Answer
A. 4Explanation
fork is an operation wherein the process makes another copy of itself.Rate this question:
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- 8.
What is the output of this C code? void main() { int i, j; for(i=0,j=0;i<10,j<20;i++,j++){ printf("i=%d \t j=%d\n", i, j); } }
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Print i and j till 19
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Print i till 9 and j till 19
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Print i and j till 9
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Runtime error
Correct Answer
A. Print i and j till 19Explanation
The code uses a comma operator in the condition of the for loop. The comma operator evaluates both expressions and returns the value of the second expression. In this case, the loop will continue as long as both i is less than 10 and j is less than 20. Since both variables are incremented in each iteration, they will eventually reach 19. Therefore, the code will print the values of i and j until they both reach 19.Rate this question:
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- 9.
What is the output of this C code? #include<stdio.h> int main() { int x=1, y=0,z=5; int a=x&&y||++z; printf("%d",z++); }
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1
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5
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6
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7
Correct Answer
A. 6Explanation
and operator doesnt operate if second argument is zero. & operator has more precedence than ||Rate this question:
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- 10.
What is the output of this C code? #include<stdio.h> int main() { int y = 2; int z = y +(y = 10); printf("%d\n", z); }
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2
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4
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20
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Compile error
Correct Answer
A. 20Explanation
The output of this C code is 20. In the line "int z = y +(y = 10)", the value of y is first assigned as 10 and then added to itself, resulting in z being assigned a value of 20. The printf statement then prints the value of z, which is 20.Rate this question:
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Current Version
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Apr 21, 2023Quiz Edited by
ProProfs Editorial Team -
Jan 06, 2008Quiz Created by
Mohit Mishra
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