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I am new to C programming but I stumbled on this code
int print(int nb)
{
if (nb < 0)
{
return (0);
}
printf("%d", nb + print(nb - 1));
nb --;
return (nb);
}
int main(void)
{
print(4);
return (0);
}
I ran the code and it gave me an output of 00246
why is that the output that, looking at it logically, the answer is not suppose to start with a 0
print(4) -> print(3) -> print(2) -> print(1) -> print(0) -> print(-1)
print(-1) stops the recursion returning 0, thus a call to printf() is emitted with 0 + 0, which is 0.
print(0) ends with -1 as value, and a call to printf() with 1 + -1 is emitted, which is 0.
etc.
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Errors:
Test isPower2(0[0x0]) failed.
Gives 1[0x1]. Should be 0[0x0]
Code:
int isPower2(int x) {
int nonNega = (x>>31);
int result = !((x & (x-1)) ^ nonNega);
return result;
}
Your isPower2(0) returns true (1) but 0 is not a power of 2. So the expected result would be false (0).
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I am trying to understand how recursion works. I seem to not understand how this multiple recursion works. Please help and thanks!
Here is the output
#include<stdio.h>
int R(int x);
int main()
{
R(5);
return 0;
}
int R(int x)
{
if(x > 0){
x--;
R(x);
R(x - 2);
printf("%d ", x);
}
}
R(5) calls R(4) which will display something and then R(2) which will also display something. After that R(5) displays 4. Which is why you see a 4 at the end of the output.
R(4)display - R(2)display - 4
Before the 4 you have the display of R(2). R(2) calls R(1) which displays somenthing and then R(0) which displays nothing. R(2) displays 1. Which is why you have a 1 at the end, just before 4.
R(4)display - R(1)display - R(0)display - 1 4
And so on...
R(4)display - R(1)display - 1 4
R(4)display - 0 1 4
You always process the R()display calls starting from the end (right side first).
As you can see recursivity will inverse the order of the numbers (4, the biggest number, is last). This is because you have the printf after the recursive calls to R.
Take a look at the trace tree I made
R(5) means the call of R function for 5, and as you know the output of it will be 4. The numbers show in which order the functions are executed till the end. The ones that have a tick besides them will make outputs.
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What does my program not output anything. I have tried other similar versions of solutions for project euler problem 1. I don't need the answer I would just like to know why their is no output. After I compile with gcc and execute the file it seems like is freezes with no output. I have to ctrl-z to kill the program.
#include <stdio.h>
/* Project Euler Problem 1 */
int main()
{
int sum = 0;
int i = 0;
while (i <= 1000);
{
if (i % 3 == 0 || i % 5 == 0);
{
sum += i;
}
i++;
}
printf("%d\n", sum);
return 0;
}
You are closing the while without doing any instruction by putting a semi-colon:
while (i <= 1000);
You should drop the semicolon.
The same for the if instruction:
if (i % 3 == 0 || i % 5 == 0);
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Please excuse me for asking this, because I know the code I'm gonna give you is wrong. Being a newbie I am not able to find the fault. Please help me correct the question and give a solution as well. Again I'm sorry to bother with this simple problem. Tomorrow is my exm in C so i'm kinda desperate. :(
Q: What will be the output of the program?
First let me show you how I find the code first:
#include<stdio.h>
int funct l(int n){
if (n>3)
return int funct(n-3)));
}
main() {
int n= 10;
printf("%d", funct l (n));
}
Then I thought i'd correct it. Then I cleaned up the code as far as i can. Then the code came to this:
#include<stdio.h>
int funct(int n){
if (n>3){
return funct(n-3);
}
}
main() {
int n= 10;
printf("%d", funct(n));
}
still it doesn't give proper answer (though I don't know what it'll show). It is either 1 or 2 and process returned 1 (0*1) is showing at the last line.
Please help me out!
Your funct function doesn't always return a value. This means that it could return anything. Try this:
int funct(int n) {
if (n > 3)
return funct(n - 3);
return n;
}
Here is the call stack when n = 10
funct(n = 10)
funct(n = 7)
funct(n = 4)
funct(n = 1)
return 1
return 1
return 1
return 1
Here is the call stack when n = 11
funct(n = 11)
funct(n = 8)
funct(n = 5)
funct(n = 2)
return 2
return 2
return 2
return 2
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Recently I am reading Mastering Algorithms with c, and in this book I have 1 exercise that I am not able to implement with c.
Tn = 1 if n=1 ,
Tn = 2T(n/2) + n if n > 1
Anyone can help me? I'll appreciate it a lot.
I have tried.
#include <stdio.h>
int test(int n) {
if (n == 1)
return 1;
else if( n > 1 )
return test(n / 2) * 2 + n;
}
int testtail(int n, int running_result) {
if (n == 1)
return running_result;
else
**return testtail(n / 2, ???? );** // How can I implement the second param
}
I am sorry guys! I am not a native English speaker! Maybe I made some mistakes in grammer! I should apologize for this!