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What happens if you try to compile and run this program?
#include
int main (int argc, char *argv[]) {
int i = 20;
printf("%x", i);
return 0;
}
-
Choose the right answer:
Answer : B
The program outputs 14 because the printf function prints the value of i as a hexadecimal integer using the %x format specifier. The hexadecimal system uses 16 symbols to represent numbers, from 0 to 9 and from A to F. Each symbol corresponds to a decimal value, for example, A is 10, B is 11, C is 12, and so on. To convert a decimal number to a hexadecimal number, we need to divide the number by 16 repeatedly and write down the remainder in reverse order. For example, to convert 20 to hexa-decimal, we do:
20 / 16 = 1, remainder 4 1 / 16 = 0, remainder 1
The hexadecimal number is 14, as we write the remainders from right to left.
Reference = CLA -- C Certified Associate Programmer Certification, C Essentials 2 - (Intermediate), C printf and scanf functions, Hexadecimal number system
What happens if you try to compile and run this program?
#include
int main (int argc, char *argv[]) {
int main, Main, mAIN = 1;
Main = main = mAIN += 1;
printf ("%d", MaIn) ;
return 0;
}
Choose the right answer:
Answer : C
The program is not a valid C program and cannot be compiled successfully. The reason is that the program uses the same name main for both a function and a variable, which is not allowed in C. The name main is a reserved keyword that denotes the entry point of the program, and it cannot be redefined or reused for any other purpose. Therefore, the compiler will report an error and the program will not run. Reference = C - main() function - Tutorialspoint, C Keywords - GeeksforGeeks, C Basic Syntax
What happens if you try to compile and run this program?
#include
int main (int argc, char *argv[]) {
char *p = "John" " " "Bean";
printf("[%s]", p) ;
return 0;
}
Choose the right answer:
Answer : C
The string literal 'John' ' ' 'Bean' is effectively concatenated into a single string by the compiler during compilation. Therefore, the value of p becomes a pointer to the string 'John Bean'. The printf statement then prints the string enclosed within square brackets, resulting in the output [John Bean].
What happens if you try to compile and run this program?
#include
int i = 0;
int main (int argc, char *argv[]) {
for(i; 1; i++);
printf("%d", i);
return 0;
}
Choose the right answer:
Answer : C
The for loop in the program is initialized with i (which is 0), has the condition 1 (which is always true), and increments i in each iteration. Since the loop con-dition is always true, the loop will continue indefinitely, and i will keep incre-menting. The program will not reach the printf statement, and it will be stuck in an infinite loop.
* The program defines a global variable i and assigns it the value 0.
* The program defines a main function that takes two parameters: argc and argv.
* The program uses a for loop to increment the value of i as long as the condi-tion 1 is true, which is always the case.
* The program never exits the for loop, so it never reaches the printf function or the return statement.
* The program keeps running indefinitely, consuming CPU resources and memory. This is an example of a logical error in the program.
What happens if you try to compile and run this program?
#include
int main (int argc, char *argv[]) {
int i = 1;
for(;i > 128;i *= 2);
printf("%d", i) ;
return 0;
}
-
Choose the right answer:
Answer : C
The main function declares an integer i and initializes it to 1. Then, it enters a for loop with no initialization statement, a condition i > 128, and an iteration expression i *= 2. The loop will continue to execute as long as i is greater than 128, but since i starts at 1, the loop's condition is false right from the start, meaning the loop body never executes.
However, this looks like an oversight, because usually, with this kind of loop, the intention is to run the loop until the condition becomes false. If the condition were i < 128, i would double each iteration until it reached or exceeded 128.
Given the current condition i > 128, the loop does nothing, and printf will output the ini-tial value of i, which is 1.
