# include<stdio.h>
int somthing(int);
int main()
{
int i=9;
somthing(i);
return 0;
}
int somthing(int i)
{
if(i == 0)
return 0;
else
printf("%d,", i);
somthing (i--);
}
I am not getting why compiler prints only 9 9 9 and then says time out that is its running infinitely.
I did post decrement on i variable one less value should go into loop every time and it should print values from 9 to 0 , code works fine when doing pre decrement on i variable .
I did post decrement on i variable one less value should go into loop every time and it should print values from 9 to 0
what you are using i-- is post increment. It would send the value of i to the something() function and then decrement the value of i. so, the somthing() function always gets the value 9. thus it goes to an infinite recursive call.
what you need is preincrement --i. It will decrement the value of i and then send it to the function somthing()
int somthing(int i)
{
if(i == 0)
return 0;
else
printf("%d,", i);
somthing (--i);
}
Do this instead. See if this works
Related
In this scenario, why is it that the output for i is 1 and not 0, since the while loop decrements the value for i two times from it's original value i=2?
#include<stdio.h>
int main(){
int i=2,j=2;
while(i+1?--i:j++)
printf("%d",i);
return 0;
}
It is because your complicated ternary operator can be reduced to
while(--i)
Explanation.
i+1 == 3, --i executes. i == 1 so the body of while loop executes (prints 1). now i+1 == 2, --i executes, i == 0 which is false in C and body of the while loop is being skipped.
i is decremneted twice - but printf executed only 1 time.
You can test it adding one more printf:
int main(){
int i=2,j=2;
while(i+1?--i:j++)
printf("%d\n",i);
printf("%d\n",i);
return 0;
}
and the result will be as expected
1
0
I'm currently learning C and I'm facing a problem that I can't solve.
It's really simple, I want the numbers from 1 to 20 printed but only if they're even (meaning the output would be 2, 4, 6,...). I know how to do this but the problem is that I want to do it in a way that I put two conditions in a for loop:
int main() {
for (int i = 1; i<=20 && i%2 == 0 ; i++) {
printf("%d\n", i);
}
return 0;
}
It seems that the second condition isn't evaluated, moreover it causes the program not to print anything. How can I do it ?
The second condition is evaluated. It is because it is evaluated the loop does not iterate.
The initial value of i is equal to 1. So the sub-expression i%2 == 0 evaluates to false.
You could use an if statement within the loop like
for (int i = 1; i<=20; i++) {
if ( i % 2 == 0 ) printf("%d\n", i);
}
If you want to place the expression i % 2 == 0 in the condition of the loop then the loop can look for example the following way as it is shown in the demonstrative program
#include <stdio.h>
int main(void)
{
for ( int i = 1; ( i % 2 == 0 ? i : ++i ) <= 20; i++) {
printf("%d\n", i);
}
return 0;
}
The program output is
2
4
6
8
10
12
14
16
18
20
Both conditions are evaluated - that's why the loop never gets going, in fact. When for starts, it sets i to 1, and immediately does the loop termination check: 1<=20 && 1%2==0. This reduces to true && false and finally to false. So the for loop won't do anything else: the very first check fails.
Remember: a failure of the condition check in the for loop terminates the loop!
Instead, you need to write what you meant - it'll be more readable and will work, too! You said:
I want the numbers from 1 to 20 printed but only if they're even
That translates directly to C (the return is unnecessary):
#include <stdio.h>
int main() {
// for all numbers from 1 to 20
for (int i = 1; i<=20; i++) {
// print out only those that are even
if ((i%2) == 0) printf("%d\n", i);
}
}
You can try this out online!.
The loop stops as soon as the condition fails. Since i%2 == 0 fails for i = 1, the loop stops immediately.
You need to put the even check inside the loop, not in the for condition.
int main() {
for (int i = 1; i<=20; i++) {
if (i % 2 == 0) {
printf("%d\n", i);
}
}
return 0;
}
I've used Code::Blocks to find the result, and it gives me 2 at the end, where i + j = 1 + 1.
#include <stdio.h>
int main(void) {
int i = -1, j = 1;
for(i++; i++; i++)
j++;
printf("%d",i + j);
return 0;
}
How does i get to be 1, and why was j not incremented?
You start with i == -1, j == 1.
Then you run the first i++, which will make i == 0. The result (which is the original value of i) is unused.
Then you test for the first time in i++. Since i is already 0, the test is false, so the for loop will quit. Now the post-increment runs, which makes it i == 1, and the loop exits.
Note how j++; was never reached!
To see it more clearly, you can think about the for loop in the following equivalent way:
i++; // init statement
while (i++) { // condition
j++; // body of the for loop
i++; // iteration expression
}
Here's the order in which it executes:
i is -1 and j is 1
i is incremented (after which i == 0)
The loop checks if i != 0. Since i is 0 at this point, the contents of the loop are skipped.
i is incremented again (after which i == 1)
The code prints i + j, which is 2 because j is unchanged and i is 1.
Here's a program with a while loop that does the same thing as the program in your question:
int main(void) {
int i = -1, j = 1;
i++;
while(i++)
{
j++;
i++;
}
printf("%d",i + j);
return 0;
}
To directly answer your question:
i is 1 afterwards because it is incremented twice from an original value of -1.
j is not incremented because the contents of the loop are skipped.
Let me first explain how for loop execution happens:
The init step is executed first, and only once. This step allows you to declare and initialize any loop control variables. You are not required to put a statement here, as long as a semicolon appears.
Next, the condition is evaluated. If it is true, the body of the loop is executed. If it is false, the body of the loop does not execute and the flow of control jumps to the next statement just after the for loop.
Increment operator: value of i increments after execution if it is i++
Explanation of your code:
for(i++; i++; i++)
At init step i++ been executed but i will be still 0 until it get to condition step of for loop.
i becomes zero it means condition is false but at same time one more i++ already executed.
As condition is false for loop statement will never be executed and value of i will be 1 and value of j remains 1.
#include< stdio.h>
int main()
{
int n,a=0,i=1,b=11;
while(i<=2)
{
while(i>0)
{
a=a+b;
i--;
}
printf("%d",a);
i++;
}
}
but if i make a little change i will get the output..
what is the difference between both the code??
#include< stdio.h>
int main()
{
int n,a=0,i=1,b=11;
while(i<=2)
{
n=i;
while(n>0)
{
a=a+b;
n--;
}
printf("%d\n",a);
i++;
}
}
output-
11
33
while (i <= 2)
{
while (i > 0)
{
a = a + b;
i--; <- out the inner while loop when i = 0
}
printf("%d", a);
i++; <- at here, the i==0 each time, so infinity loop
}
Because your nested loop always restores the value of i to 0,
And 0 <= 2 is always true, thus it keeps on going.
Initially the value of i is 1, the first loop starts by checking i<=2 , which is true for i=1, then the second loop checks i>0, which is also true, then the second loop decreases the value of i to 0 by i--;
This time the test condition for the nested loop fails and the inner loop exits, back in the first loop, the condition is satisfied as i=0 is<= 2 thus i is incremented, now i = 1.
This keeps on going forever.
In the second code, you're copying the current value of i to n, thus, initially i = 1;
Condition for first loop satisfied, then you set n = i,
And check n >0 which is true as n = i = 1
In this loop you are decrementing n and n becomes 0, thus the loop quits, and the outer loop increments i, i now being 2 allows the outer loop to run and then again n = i = 2,
The inner loop runs twice and then exits and i is incremented to 3 failing the outer loop condition and hence quitting the loop. And you get the result.
In the first code your
While(i<=2) never ends because you add 1 to i with i++ and then you subtract 1 from i with i-- and you never get i=3 to end the while loop.
I'm trying to understand my professor's code for teaching permutations, and I don't know what the "(!used[i])" inside an if statement means or does. Here's the full function, the if statement is within the for loop. Can anyone explain what it does?
void RecursivePermute(int n, int k, int* perm, int* used) {
int i;
// We've filled perm already. Print the corresponding permutation.
if (k == n) {
for (i=0; i<n; i++)
printf("%d ", perm[i]);
printf("\n");
}
// Try each unused number in spot k.
for (i=0; i<n; i++) {
if (!used[i]) { //this if statement is my question
perm[k] = i;
used[i] = 1;
RecursivePermute(n, k+1, perm, used);
used[i] = 0;
}
}
}
It means not, so will trigger the if statement when the int element used[i] == 0, so it could also be written as:
if (used[i] == 0) {
...
}
Truth statements treat a variable as false if it's 0 or true if it's anything else. ! is the not operator. Therefore !used[i] will do the opposite, return true if used[i] is zero, or false otherwise.
used is an array of integers. doing
if (!used[i])
checks if the current element is 0
what the recursive function do is each time print the array then put 1 in the next element, causing perm(and the output) to look like, depending on initiative k:
00000
01000
01200
01230
.
.
.