C Programming - Control Instructions - Discussion

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

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

@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.

char is one of integral type when we assign an integer value into a char then it is treated as like integer.so the ans is as like below
step1 while(j<5) 1<5 yes value y=1 print
step2 j=2 2<5 yes value y=2 print
step3 j=3 3<5 yer value y=3 print
step4 j=4 4<5 yes value y=4 print
step5 j=5 5<5 no come out of loop so the values are 1,2,3,4

int main()
{
int i=0;
for(; i<=5; i++);
printf("%d,", i);
return 0;
}

Answer for above question is 6

char j=1;
while(j < 5)
{
printf("%d, ", j);
j = j+1;
}
printf("\n");
return 0;

Answer for ths is 1 2 3 4. In first program why we are not mentioning 12345 ?

in the program j=1 but not j='1';
if j='1' then ascii value of 1 is assigned to j and while loop executes.
in the above program we just convert char to int using format strings in printf statement it is possible in c......

j=j+1 is same as j++ post increment so
If(1<5)
printf("%d,",j);
j++;
It keep on increasing unless it encounters the condition 4<5 when it become 5<5 it fails then ans is 1,2,3,4

In C char types are treated as int by the compiler. Hence they can be used in place of int (<255 for unsigned). This is the reason why char can be used inside switch statement.

here j is char which is of 1 byte. In 1 byte we can have number from 0 to 255.(2^8=256)as 1 byte=8 bits.

Hence using char as data type we can store integer values up to 255.

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

can anyone explain?