C Programming - Control Instructions - Discussion

This above statement will print value along with comma. For example:

int j=9;
printf ("%d, ", j) ;

Output:9,

Hi!

j is character data type and its range is 0 to 255 and all these are included as integer data type. So, the character data type is treated as integer data type also. And it evalutes same as int data type.
Step1:
while(j < 5) is evalutes as while(1 < 5) is true.
the printf statement executes 1
then j = j+1 evalutes to j =2 Until the while statement becomes fails.
When j = 5 , while statement fails.
The result is
1 2 3 4 only

While loop execute 4 times so will print 1, 2, 3, 4.

printf ("%d, ", j) ;.

Why there is a comma after the symbol d?

@Omkumar.

In first programme you put a semicolon right after the for statement.

for (;i<=5;i++) ;

This means that for loop doesn't have any statements to run.

So even if the for loop condition is TRUE nothing is gonna happen.

For loop will run without printing anything until the condition is TRUE. Once the condition is FALSE (i.e., When i=6) it will come out of the for loop and print i.

Thanks.

Write the same program in c++. It will give you different result. It will show the ascii value of corresponding digit.

can anyone explain?

Here, it will be int i=1.

while(1<5) it is true enters the loop
prints j values as 1
j value is incremented by 1 i.e now j value is 2
while(2<5) it is true enters the loop
prints j values as 2
j value is incremented by 1 i.e now j value is 3
while(3<5) it is true enters the loop
prints j values as 3
j value is incremented by 1 i.e now j value is 4
while(4<5) it is true enters the loop
prints j values as 4
j value is incremented by 1 i.e now j value is 5
while(5<5) it is false exits from loop

How can be char j =1?
1 is an integer so j should be int datatype.

Anyone, explain the meaning of %d in this printf("%d, ", j).