This program is supposed to asks the user to enter a positive integer (the integer could be of any number of digits in the range of the integer type) and replace each digit by the sum of that digit plus 6 modulus 10. The program then should swap the first digit with the last digit before it displays the output.
A sample input/output:
Enter the number of digits of the number: 5
Enter the number: 92828
Output: 48485
For some reason with my code, no matter what number I enter, everything just comes out as 6. (so if I enter 5 numbers, I get 666666). I'm new to pointers, so is there an issue with that, or do I just have some math wrong? The program runs without any compiler warnings.
#include <stdio.h>
#include <stdlib.h>
void replace(int *a, int *b, int n);
void swap(int *p, int *q);
int main()
{
int n = 0;
int i = 0;
int a[100], b[100];
//Prompt user to enter number of digits
printf("Enter the number of digits you'd like to replace: ");
scanf("%d", &n);
//Prompt user to enter the number to use
printf("Enter the number to use: ");
for(i = 0; i < n; i++);
scanf("%1d", &a[i]);
//replace function
replace(a, b, n);
for(i = 0; i < n; i++)
printf("%d", b[i]);
printf("\n\n");
return 0;
}
void replace(int *a, int *b, int n)
{
int i;
for (i = 0; i < n; i++)
{
*(b+i) = (*(a+i)+ 6) % 10;
}
printf("The output is: ");
//swap function
swap(b, (b+ (n-1)));
}
void swap(int *p, int *q)
{
int t;
t = *p;
*p = *q;
*q = t;
}
Your code is absolutely correct except a silly mistake in the following code snippet.
for(i = 0; i < n; i++);
scanf("%1d", &a[i]);
Why did you put a ; after the for statement? It means your for loop is just iterating once (instead of 5 if n = 5). As a result, only the first digit input is considered given by the user but that too be stored in a[5] (considering n = 5), values stored in a[0] to a[4] are all garbage value.
Just remove the semicolon and update your code as follows.
for(i = 0; i < n; i++)
scanf("%1d", &a[i]);
Now it works fine.
The culprit in your code is the semicolon after the for loop:
for(i = 0; i < n; i++)**;**
scanf("%1d", &a[i]);
So the scanf that you wrote is basically out of the for-loop and stores the first digit into a[n].
Related
I was given an assignment to write a code which takes in numbers as input from the user and provides the sum of it, specifically by the use of pointer arithmetic i.e. no array subscripting a[i] is allowed.
Below is the code that I wrote, which got compiled and even ran. But almost always it gives the sum of the input numbers as 0. I tried to fix it, but to no avail. Thus, I am asking for help, any help is greatly appreciated.
#include<stdio.h>
#define N 5
int sum_array( const int *p, int n)
{
int sum, a[N];
sum = 0;
for(p=&a[0]; p<&a[N]; p++)
sum += *p;
return sum;
}
int main()
{
int a[N], *i,x;
printf("Enter %d Numbers: ", N);
for(i=a; i<a+N; i++)
scanf("%d", i);
// all the input values get scanned as i or the array a
x= sum_array(i,N);
printf("the sum is %d\n", x);
return 0;
}
Beware, you are declaring array int a[N] in both main and sum_array. They are in different scopes, so they are different arrays (and the one from sum_array is never initialized so reading it invokes Undefined Behaviour).
The correct way is to pass the array along with its used length:
Here is a fixed version:
#include<stdio.h>
#define N 5
int sum_array( const int *a, int n) // a points to a array of at least n elements
{
int sum = 0; // initialize at definition time
for(const int *p=a; p<&a[n]; p++) // have the pointer p take all values from a
sum += *p;
return sum;
}
int main()
{
int a[N], *i,x;
printf("Enter %d Numbers: ", N);
for(i=a; i<a+N; i++)
scanf("%d", i);
// all the input values get scanned as i or the array a
x= sum_array(a,N); // pass the array address, not a pointer past last element
printf("the sum is %d\n", x);
return 0;
}
Finally it is mainly a matter of taste, but I was too often burnt for trying to add an instruction in a for loop without braces, so I strongly recommend using always braces for loops:
for(i=a; i<a+N; i++) {
scanf("%d", i);
}
int sum_array( const int *p, int n)
{
int sum = 0, i = 0;
for(i = 0; i < n ; i++)
sum += *(p+i);
return sum;
}
int main(void)
{
int a[N], i = 0, x = 0;
printf("Enter %d Numbers: ", N);
for(i=0; i<N; i++)
scanf("%d", a+i);
// all the input values get scanned as i or the array a
x= sum_array(a,N);
printf("the sum is %d\n", x);
return 0;
}
In x= sum_array(i,N); i is the iterator of your loop so after the loop has finished it points to the first position after the array.
You should pass the original array instead x= sum_array(a,N);
In the sum function you just throw away the passed pointer and replace it with your local a[].
int sum_array( const int *p, int n)
{
int sum = 0;
int *end = &p[n]; // first element after the array.
for(; p<end; p++) // just use p because you don't need the reference to the start of the array
{
sum += *p;
}
return sum;
}
but as you said that array notation is not allowed you can change it as follows
#include "stdio.h"
#define N 5
int sum_array( const int *p, int n)
{
int sum = 0;
const int *end = p+n; // first element after the array.
for(; p<end; p++)
{
sum += *p;
}
return sum;
}
int main()
{
int *a, *i, x;
a = malloc(N * sizeof(*a));
if (a == NULL)
exit(-1);
printf("Enter %d Numbers: ", N);
for(i=a; i<a+N; i++)
{
scanf("%d", i);
}
// all the input values get scanned as i or the array a
x= sum_array(a,N); // pass the array address, not a pointer past last element
printf("the sum is %d\n", x);
return 0;
}
in general, when programming, the code should be kept as simple as possible while still being complete.
The program criteria shows no need to keep a number after it is applied to the sum of the numbers, So in the proposed code, the input number is only kept long enough to be applied to the sum, then it is discarded.
The following proposed code:
cleanly compiles
performs the desired functionality
is kept very simple
properly checks for; and handles any errors
And now the proposed code:
#include <stdio.h> // scanf(), printf(), fprintf(), stderr
#include <stdlib.h> // exit(), EXIT_FAILURE
#define N 5
int main( void )
{
int num = 0;
int sum = 0;
printf("Enter %d Numbers: ", N);
for(size_t i=0; i<N; i++)
{
if( scanf("%d", &num) != 1 )
{
fprintf( stderr, "failed to read number\n" );
exit( EXIT_FAILURE );
}
// implied else, scanf successful
sum += num;
}
printf( "the sum is %d\n", sum );
return 0;
}
I've been trying to apply all advices found in this site but none seems to be working.
For the first part of the code I need to fill an array with random numbers (0 or 1) to simulate an epidemic spreading, but the array obtained is not the desired one at all... this is the code I wrote:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main(int argc, char **argv)
{
int N, BC, t, T, i, v[N];
float b, g, p, r;
/*Variable values initialization*/
printf("Enter infection probability:\n");
scanf("%f", &b);
printf("Enter the number of individuals:\n");
scanf("%d", &N);
printf("Enter the number of time steps:\n");
scanf("%d", &T);
printf("Periodic boundary contitions? (Y:1 / N:0)\n");
scanf("%d", &BC);
/*First set of individuals*/
srand(time(NULL));
for(i = 0; i < N; i++){
v[i] = (rand()/RAND_MAX);
}
/*Check if array properly initialized*/
printf("Initial array:\n" );
for(i = 0; i < N; i++){
printf("%d-", v[i]);
}
The outcome I expected for the array was something like: 1-0-1-1-0-0-0-..., but I always get the following one:
Initial array:
0-0-2-15-0-0-0-0-0-0-
What am I doing wrong?
Thanks a million!
You should declare v[N] after
printf("Enter the number of individuals:\n");
scanf("%d", &N);
otherwise its size will be random since N isn't initialized when the memory allocated for v[] based on N is set.
If you want just 0 or 1 you should use a modulo:
srand(time(NULL));
for(i = 0; i < N; i++){
v[i] = (rand() % 2);
}
all the even values generated by rand will become 0 and all the odd values will become 1
I want to do this code that tells you the number of (n) integers that are bigger (or equal) than a (k) input.
So for example:
input:
4 15
12
6
15
24
output:
2
So the 4 is the number of integers the user is going to input and the 15 is the k number, now the output is the quantity of numbers that are bigger than k.
What I have of code is this:
#include<stdio.h>
int main()
{
int n, k, i;
int c, d;
scanf(" %d",&n);
scanf("%d", &k);
for(i=1;i<=n;i++)
{
scanf("%d",&c);
if (c[i]>k)
c[i]=d;
}
printf("%d", d);
return 0;
}
As you can see my code sucks, I don't know how to find the integers that are bigger than k and to print the quantity, not the numbers themselves. Any help is really appreaciated. Thanks.
Far less elegant solution, but one that keeps the value you need for some further use.. OldProgrammer did it much simpler and more pretty.
int main()
{
int num, s, i, cnt = 0;
printf("please input number of integers and int to compare with\n");
scanf("%d %d", &s, &num);
int arr[s];
for(i = 0; i < s; i++)
{
printf("Please input %d. number", i+1);
scanf("%d", &arr[i]);
}
for(i = 0; i < s; i++)
{
if(arr[i] >= num)
cnt++;
}
//at this point cnt holds the value you need
return 0;
}
Not sure why you are trying to reference c as an array. That is not needed. Try this:
int main()
{
int n, k, i, c;
int count = 0;
scanf(" %d",&n);
scanf("%d", &k);
for(i=1;i<=n;i++)
{
scanf("%d",&c);
if (c > k)
count++;
}
printf("%d", count);
return 0
}
Also, I would rename your variables to something more meaningful, such as numEntries, checkValue, etc.
In my program, after I enter the first input. I expected the values to get printed because of the printf statement next to scanf, but for the first time alone it's not getting printed.For the next iterations and all its working correctly
Ex:
Output :
Enter input10
12
i=0,a[i]=11896224
Enter input
Code:
#include<stdio.h>
#include<stdlib.h>
int func(int *, int);
int *a;
int main()
{
int length = 5;
a = (int *)malloc(sizeof(int)*length);
for (int i = 0;i < length;i++)
{
printf("Enter input");
scanf("%d\n", &a[i]);//100,104,108,112,116
printf("i=%d,a[i]=%d\n", i, &a[i]);/*Print Statement which is not executing for first time*/
}
func(a, length);
return 0;
}
int func(int *b, int length)
{
printf("Length=%d", length);
for (int j = 0;j < length;j++)
printf("b[%d]=%d", j, b[j]);//
return 0;
}
Remove the \n from the scanf function.
Calculating average three by three elements and replacing those elements with the average result.
Example array [1,2,7,-2,5,0, 2,8]
After transformation [3,3,3,1,1,1,5,5]
Something is wrong, I can't get it to work.
#include <stdio.h>
int main ( ) {
int n, c[n];
int *avg;
int pom=0;
printf("Enter lenght of array\n");
scanf("%d",&n);
printf("Enter elements");
for(i = 0;i < n; i++)
scanf("%d",c[i]);
avg=Average(c , n, pom);
for(i = 0; i < n; i++)
printf("Avg elements= %d",*(avg+i))
return 0;
}
int Average(int arr[], int size, int z)
{
int k, l, m, Asum;
if (size < 0) {
return arr;
} else {
k=arr[z];
l=arr[z+1];
m=arr[z+2];
Asum=(k + l + m)/3;
arr[z]=Asum;
arr[z+1]=Asum;
arr[z+2]=Asum;
}
return Average(arr,size--,z++);
}
int n, c[n]; is a problem. n is uninitialized so the size of the array is who-knows-what? This is undefined behavior.
Instead
int main(void) {
int n;
int *avg;
int pom=0;
printf("Enter length of array\n");
if (scanf("%d",&n) != 1) return -1;
int c[n];
for(i = 0;i < n; i++)
// scanf("%d",c[i]);
scanf("%d",&c[i]); // pass the address of an `int`
Likely other issues too.
Try simple input first, imagine what happens when you enter only 1 number, what will the Average function do? Don't run the code but try to execute it in your head or with pencil and paper. If you think the program only has to work with three or more numbers, try three.
A serious program would explicitly reject invalid input.