C Exercises: Check whether an integer is a palindrome or not
C Programming Mathematics: Exercise-2 with Solution
Write a C program to check whether an integer is a palindrome or not.
An integer is a palindrome when it reads the same forward as backward.
Example:
Input:
i = 1221
i = -121
i = 100
Output:
Is Palindrome: 1
Is Palindrome: 0
Is Palindrome: 0
Pictorial Presentation:
Sample Solution:
C Code:
#include <stdio.h>
#include <stdbool.h>
bool is_Palindrome(int i) {
int n, d, y = 0;
if (i < 0)
return false;
n = i;
while (n) {
d = n % 10;
if (y > (0x7fffffff - d) / 10)
return false;
y = y * 10 + d;
n = n / 10;
}
return (y == i);
}
int main(void)
{
int i = 1221;
printf("Original integer: %d ",i);
printf("\nIs Palindrome: %d ",is_Palindrome(i));
i = -121;
printf("\nOriginal integer: %d ",i);
printf("\nIs Palindrome: %d ",is_Palindrome(i));
i = 100;
printf("\nOriginal integer: %d ",i);
printf("\nIs Palindrome: %d ",is_Palindrome(i));
return 0;
}
Sample Output:
Original integer: 1221 Is Palindrome: 1 Original integer: -121 Is Palindrome: 0 Original integer: 100 Is Palindrome: 0
Flowchart:
C Programming Code Editor:
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Next: Write a C program to divide two integers (dividend and divisor) without using multiplication, division and mod operator.
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C Programming: Tips of the Day
Static variable inside of a function in C
The scope of variable is where the variable name can be seen. Here, x is visible only inside function foo().
The lifetime of a variable is the period over which it exists. If x were defined without the keyword static, the lifetime would be from the entry into foo() to the return from foo(); so it would be re-initialized to 5 on every call.
The keyword static acts to extend the lifetime of a variable to the lifetime of the programme; e.g. initialization occurs once and once only and then the variable retains its value - whatever it has come to be - over all future calls to foo().
Ref : https://bit.ly/3fOq7XP
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