We are coding to count the number of lines and the corresponding numbers.
For example, if you enter 5, the calculation is
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
The coding that results in this way.
First, you specified the count variable as cnt.
However, if you specify cnt++; outside the printf sentence and printf("%d",cnt++), the result will be different.
I wonder why.
Below is the coding I did.
int n,cnt=1;
scanf_s("%d", &n);
for (int a = 1; a <= n; a++) {
for (int b = 1; b <= a; b++) {
printf("%d",cnt++); //If you insert cnt++ here, it should be initialized to cnt=1.
} // If you just put cnt++; above, it is cnt=0.
printf("\n");
}
But here,
cnt++;
printf("%d", cnt)
in the case of You need to initialize cnt to 0 to get the above results.
As above
printf("%d ",cnt++);
In this way, cnt should be initialized to 1 to get the above result.
I wonder how there is a difference of 1.
c
The ++ operator increases the variable value by 1 when used alone, but when used in other sentences, it affects the order of evaluation of sentences.
The ++ operator can be put before or after a variable, and the order of evaluation varies.
printf("%d ",cnt++);
One line of code above is treated the same as the two lines of code below in a typical compilation environment.
printf("%d ",cnt);
cnt++;
That is, after outputting the current cnt value on the screen, the cnt value increases by 1.
In the question, the printf("%d",cnt++);code also increased after the current cnt value was output, so you had to initialize cnt to 1 to output from 1.
On the other hand, in the question
cnt++;
printf("%d", cnt);
In this case, the cnt value should be increased by 1 and then the cnt should be output from 1, so the output value should be increased by 1 and then the output value should be 1, so the cnt had to be initialized to 0.
For your information,
printf("%d ",cnt++);
To avoid misunderstanding rather than writing like this, the two-line code below is better written.
printf("%d ",cnt);
cnt++;
© 2024 OneMinuteCode. All rights reserved.