C Programming - Control Instructions
#include<stdio.h>
int main()
{
int i = 10, j = 20;
if(i = 5) && if(j = 10)
printf("Have a nice day");
return 0;
}
"Expression syntax" error occur in this line if(i = 5) && if(j = 10).
It should be like if((i == 5) && (j == 10)).
#include<stdio.h>
int main()
{
int i = 10, j = 15;
if(i % 2 = j % 3)
printf("IndiaBIX\n");
return 0;
}
if(i % 2 = j % 3) This statement generates "LValue required error". There is no variable on the left side of the expression to assign (j % 3).
#include<stdio.h>
int main()
{
int x = 30, y = 40;
if(x == y)
printf("x is equal to y\n");
else if(x > y)
printf("x is greater than y\n");
else if(x < y)
printf("x is less than y\n")
return 0;
}
This program will result in error "Statement missing ;"
printf("x is less than y\n") here ; should be added to the end of this statement.
1: | Every if-else statement can be replaced by an equivalent statements using ?: operators |
2: | Nested if-else statements are allowed. |
3: | Multiple statements in an if block are allowed. |
4: | Multiple statements in an else block are allowed. |
#include<stdio.h>
int main()
{
int i = 0;
i++;
if(i <= 5)
{
printf("IndiaBIX\n");
exit(0);
main();
}
return 0;
}
Step 1: int i = 0; here variable i is declared as an integer type and initialized to '0'(zero).
Step 2: i++; here variable i is increemented by 1(one). Hence, i = 1
Step 3: if(i <= 5) becomes if(1 <= 5) here we are checking '1' is less than or equal to '5'. Hence the if condition is satisfied.
Step 4: printf("IndiaBIX\n"); It prints "IndiaBIX"
Step 5: exit(); terminates the program execution.
Hence the output is "IndiaBIX".