Question 1 |
Predict the output of following program?
#include "stdio.h" int main() { char arr[100]; printf("%d", scanf("%s", arr)); /* Suppose that input value given for above scanf is "GeeksQuiz" */ return 1; }
9 | |
1 | |
10 | |
100 |
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Question 1 Explanation:
In C, scanf returns the no. of inputs it has successfully read. See https://tutorialspoint.dev/slugresolver/archives/674
Question 2 |
Predict output of the following program
#include <stdio.h> int main() { printf("new_c_questionby"); printf("rgeeksforgeeks"); getchar(); return 0; }
ew_c_question geeksforgeeks | |
new_c_ques geeksforgeeks | |
geeksforgeeks | |
Depends on terminal configuration |
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Question 2 Explanation:
Question 3 |
#include <stdio.h> int main() { printf(" "GEEKS %% FOR %% GEEKS""); getchar(); return 0; }
“GEEKS % FOR % GEEKS” | |
GEEKS % FOR % GEEKS | |
"GEEKS %% FOR %% GEEKS" | |
GEEKS %% FOR %% GEEKS |
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Question 3 Explanation:
Backslash (\\) works as escape character for double quote (“). For explanation of %%, see https://tutorialspoint.dev/slugresolver/how-to-print-using-printf/
Question 4 |
#include <stdio.h> // Assume base address of "GeeksQuiz" to be 1000 int main() { printf(5 + "GeeksQuiz"); return 0; }
GeeksQuiz | |
Quiz | |
1005 | |
Compile-time error |
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Question 4 Explanation:
printf is a library function defined under stdio.h header file. The compiler adds 5 to the base address of the string through the expression 5 + "GeeksQuiz" . Then the string "Quiz" gets passed to the standard library function as an argument.
Question 5 |
Predict the output of the below program:
#include <stdio.h> int main() { printf("%c ", 5["GeeksQuiz"]); return 0; }
Compile-time error | |
Runtime error | |
Q | |
s |
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Question 5 Explanation:
The crux of the program lies in the expression: 5["GeeksQuiz"]
This expression is broken down by the compiler as: *(5 + "GeeksQuiz"). Adding 5 to the base address of the string increments the pointer(lets say a pointer was pointing to the start(G) of the string initially) to point to Q. Applying value-of operator gives the character at the location pointed to by the pointer i.e. Q.
Question 6 |
Predict the output of below program:
#include <stdio.h> int main() { printf("%c ", "GeeksQuiz"[5]); return 0; }
Compile-time error | |
Runtime error | |
Q | |
s |
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Question 6 Explanation:
The crux of the program lies in the expression: "GeeksQuiz"[5].
This expression is broken down by the compiler as: *(“GeeksQuiz” + 5). Adding 5 to the base address of the string increments the pointer(lets say a pointer was pointing to the start(G) of the string initially) to point to Q. Applying value-of operator gives the character at the location pointed to by the pointer i.e. Q.
Question 7 |
Which of the following is true
gets() can read a string with newline chacters but a normal scanf() with %s can not. | |
gets() can read a string with spaces but a normal scanf() with %s can not. | |
gets() can always replace scanf() without any additional code. | |
None of the above |
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Question 7 Explanation:
gets() can read a string with spaces but a normal scanf() with %s can not. Consider following program as an example.
If we enter "Geeks Quiz" as input in below program, the program prints "Geeks"
But in the following program, if we enter "Geeks Quiz", it prints "Geeks Quiz"
Question 8 |
Which of the following is true
gets() doesn't do any array bound testing and should not be used. | |
fgets() should be used in place of gets() only for files, otherwise gets() is fine | |
gets() cannot read strings with spaces | |
None of the above |
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Question 8 Explanation:
Question 9 |
What does the following C statement mean?
scanf("%4s", str);
Read exactly 4 characters from console. | |
Read maximum 4 characters from console. | |
Read a string str in multiples of 4 | |
Nothing |
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Question 9 Explanation:
Try following program, enter GeeksQuiz, the output would be "Geek"
#include <stdio.h> int main() { char str[50] = {0}; scanf("%4s", str); printf(str); getchar(); return 0; }
Question 10 |
#include<stdio.h> int main() { char *s = "Geeks Quiz"; int n = 7; printf("%.*s", n, s); return 0; }
Geeks Quiz | |
Nothing is printed | |
Geeks Q | |
Geeks Qu |
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Question 10 Explanation:
.* means The precision is not specified in the format string, but as an additional integer value argument preceding the argument that has to be formatted.
There are 25 questions to complete.
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