#include <stdio.h>
int main()
{
int n, i;
double num[100], sum = 0.0, average;
printf("Enter the numbers of elements: ");
scanf("%d", &n);
for(i = 0; i < n; i++)
{
printf("Enter number %d: ", i+1);
scanf("%f", &num[i]);
sum += num[i];
}
average = sum / n;
printf("Average = %f", average);
return 0;
}
Hi, I have a problem with displaying the average, displaying numbers from space, how to fix it?
Correct format specifier for double would be (you will have to use this)
scanf("%lf", &num[i]);
For printf both %f or %lf will do. Note that when using scanf check the return value of scanf - in case of failure you will take necessary action. (Wrong input given or some other error).
Using the correct format specifier for double as stated by #coderredoc resolves it. Use %lf instead of %f.
Related
When I change the type of array from float to double (float array[][] to double array[][]), it doesn't scan the values correctly. All the values become zero. For example, if I enter 5 for input when it's float, it's 5.000000. However, when it's double, every value I enter is scanned as 0.0000000.
#include <stdio.h>
#include <math.h>
int main() {
int limestone, min=0;
//for entering height of array (number of limetstone)
printf("Enter number of limestone: ");
scanf("%d", &limestone);
float array[limestone][3];
//double array[limestone][3]; (the problem)
//for getting inputs
for(int i=0;i<limestone;i++)
{
printf("Enter the %d porosity, hydraulic conductivity (m/s), specific gravity: ", i+1);
scanf("%f %f %f", &array[i][0], &array[i][1], &array[i][2] );
}
//for print array (you can remove it)
for(int i=0;i<limestone;i++)
{
for(int j=0;j<3;j++)
{
printf("%f ",array[i][j]);
}
printf("\n");
}
//Comparing 3rd (Last) Column
for(int i=limestone-1;i>=0;i--)
{
if(array[i][2]<array[min][2])
{
min=i;
}
}
printf("The limestone with the lowest specific gravity is Limestone %d with a specific gravity of %f",min+1,array[min][2]);
return 0;
}
I got your code to work by changing your scanf line to
scanf("%lf %lf %lf", &array[i][0], &array[i][1], &array[i][2] );
All I did was change the %f's to %lfs.
This works for me at least, I think it's doing what you want
My array not taking inputs (Ex- 5.21).I am trying to take a float input for each of my array elements but this isn't doing this. and Also this function is not working with the array. please help
float maxJump (double jumpRecord[], int attempts)
{
float max = jumpRecord[0];
for(int i=0; i<attempts; ++i)
{
if(jumpRecord[i]>max)
max = jumpRecord[i];
}
printf("Longest Jump = %.2f meter\n",max);
}
int main()
{
double jumpRecord[5];
printf("Jumping information (in meter): \n");
for(int i=0; i<5; i++)
{
printf("Attempt %d. = ",i+1);
scanf("%.2f",&jumpRecord[i]);
}
maxJump(jumpRecord,5);
return 0;
}
When used with scanf(), %.2f (%f) format specifier is for reading float values.
The elements of jumpRecord are double, so you should use %lf (%.2lf) format specifier instead.
Note that %f should be used for printing double via printf(). %lf is allowed in C99 or later, but %f is more portable.
In my C Program, I want to get the average of the sum of numbers being entered until the program stops. What should I add to check the average? Thank you!
#include <stdio.h>
int main (void)
{
int x;
int sum = 0;
int average;
int testEOF;
//Statements
printf("Enter your numbers: <EOF> to stop.\n");
do
{
testEOF = scanf("%d", &x);
if (testEOF !=EOF)
sum +=x;
} while (testEOF !=EOF);
printf ("\nTotal: %d\n", sum);
printf ("\nAverage: %d\n", average);
return 0;
//main
}
As other people explained, you have to initialize sum and count, they are never given their initial values. No need to initialize average and x because you assign sum/count to average and users will assign any value to x.
You have to put everything you want your if statement to do in {...}. But you didn't. Your if statement only does sum +=x and does not work for count++ and average = sum / count. So your program increases the value of count even after EOF. So you found 14/6 rather than 14/5.
You checked testEOF != EOF twice. One is in the if statement, other is in do-while.
I put the code below:
#include <stdio.h>
int main (void){
float x;
float sum = 0;
float count = 0;
float average;
float testEOF;
printf("Enter your numbers: <EOF> to stop.\n");
while(1){
testEOF = scanf("%f", &x);
if (testEOF ==EOF){
break;
}
sum +=x;
count++;
average = sum / count;
}
printf ("\nTotal: %f\n", sum);
printf ("\nAverage: %.2f\n", average);
return 0;
}
#include <stdio.h>
int main (void)
{
float x;
float sum;
float count;
float average;
float testEOF;
//Statements
printf("Enter your numbers: <EOF> to stop.\n");
do
{
testEOF = scanf("%f", &x);
if (testEOF !=EOF)
sum +=x;
count++;
average = sum / count;
} while (testEOF !=EOF);
printf ("\nTotal: %f\n", sum);
printf ("\nAverage: %.2f\n", average);
return 0;
}
This is now almost correct. The only thing is that the number is not getting divided by the number of values entered.
Example:
2
3
3
3
3
Sum is 14
Average: 2.3 (should be 2.8)
This seems that count is being added by 1 every time I get the average.
Well if you want your output to be printed with decimal point you can do the following , in that case
for inputs like 0, 1, 1 the out would be 0.666667 otherwise it will be simply 0 ignoring the decimal part.
the while part can be optimized as suggested by #David C. Rankin
double sum = 0;
double average = 0;
unsigned int count = 0;
//Statements
printf("Enter your numbers: <EOF> to stop.\n");
do
{
testEOF = scanf("%d", &x);
if (testEOF != EOF)
{
sum += x;
count++;
}
} while (testEOF != EOF);
printf ("\nTotal: %f\n", sum);
average = sum / count;
printf ("\nAverage: %f\n", average);
For some reason my code will consistently print out zeros.
I was supposed to make a code in which I enter three numbers,
The first number
The ratio
The amount of numbers to be displayed
The code should display those numbers.
Here is the code:
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
void forloop(int firstNum,int ratio,int repeats);
int main(void)
{
int firstNum = 0;
int ratio = 0;
int repeats = 0;
printf("First number of the series: ");
scanf("%d", &firstNum);
printf("the ratio of the series: ");
scanf("%d", &ratio);
printf("the amount of numbers to display is ");
scanf("%d", &repeats);
forloop(firstNum,ratio,repeats);
}
void forloop(int firstNum,int ratio,int repeats)
{
int i = 0;
for (i=1; i!=repeats+1;i++)
{
printf("%d ", firstNum*(pow( ratio, i)));
}
}
This should fix the issue with printing zeros:
void forloop(int firstNum,int ratio,int repeats)
{
int i = 0;
for (i=1; i!=repeats+1;i++)
{
printf("%.0f ", firstNum*(pow((double)ratio,(double)i)));
}
}
As mentioned by #JonathanLeffler in the comments, you have a bug. The line
printf("%d ", firstNum*(pow( ratio, i)));
is wrong because pow returns a double but in the printf specifications %d expects an integer.
So, you can either change the specifier in your printf to a floating point one, i.e., %f, %g, %e, %F, %G, %E et cetera or cast pow's output to int by doing (int)pow(ratio, i).
In the first case, as #JadMrad said, if you are printing a double and you don't want to have decimal numbers, you can specify the number of digits at the right of the decimal point using "%.Nf" instead of "%f" where N is the number of decimal points that you want.
Nevertheless, in your case it seems to me that you are expecting always integer numbers. Then,
printf("%d ", firstNum*((int)pow( ratio, i)));
sounds like a better solution to me.
I wanted to create a program to calculate the regression line of some given data, along with the errors, so I can use in my university assignments. This is the following program:
#include <stdio.h>
#include <math.h>
int main(void)
{
int i,n,N;
double x[n],y[n],a=0.0,b=0.0,c=0.0,d=0.0,D,P=0.0,p,Sx,A,B,dA,dB;
printf("Give the number of data you want to input for x : ");
scanf("\n%d", &N);
printf("\nGive the values of x : ");
for (i=1; i<=N; i++);
{
printf("\n Enter x[%d] = ", i);
scanf("%lf", &x[i]);
a+=x[i];
b+=pow(x[i],2);
}
printf("\nGive the values of y : ");
for (i=1; i<=N; i++);
{
printf("\n Enter y[%d] = ", i);
scanf("%lf", &y[i]);
c+=y[i];
}
D=N*b-pow(a,2);
A=(b*c-a*d)/D;
B=(N*d-a*c)/D;
for (i=1; i<=N; i++);
{
d+=x[i]*y[i];
p=y[i]-A-B*x[i];
P+=pow(p,2);
}
Sx=sqrt(P/N-2);
dA=Sx*sqrt(b/D);
dB=Sx*sqrt(N/D);
printf("\n x \t \t \t y");
for (i=1; i<=N; i++);
printf("\nx[%d] = %lf\t%lf = y[%d]", x[i],y[i]);
printf("\nA = %lf\t B = %lf", A,B);
printf("\nThe errors of A & B are dA = %lf and dB = %lf", dA,dB);
printf("\nThe equation of the regression line is y=%lfx+(%lf)", B,A);
return 0;
}
I have two problems.
Despite giving a value to N, the program runs so that I can only give one value for x and one value for y. Why and where is the mistake?
When printing the "Enter x[%d]", it displays x[11] and at the end when printing "x[%d] = %lf\t%lf = y[%d]", it displays x[0]. Again why and where is the mistake?
Thank you for your help!
You are trying to create a dynamic array in C.
To do that, you need to use dynamic memory allocation with malloc and free. So, your code should look something like this:
int N;
double *x;
printf("Give the number of data you want to input for x :\n");
scanf("%d", &N);
x = malloc(sizeof(double) * N);
Then, at the end of your program you need to free the memory:
free(x);
If you don't want to deal with manual memory management (or can't for some reason), you can use a static maximum array size like this:
#define MAX_N_X 100
int main(void) {
int N;
double x[MAX_N_X];
printf("Give the number of data you want to input for x :\n");
scanf("%d", &N);
if (N > MAX_N_X) {
printf("Can't handle that many inputs! Maximum %d\n", MAX_N_X);
return 0;
}
}
You simply missed two parameters to printf.
Yo wrote:
printf("\nx[%d] = %lf\t%lf = y[%d]", x[i],y[i]);
But it should be:
printf("\nx[%d] = %lf\t%lf = y[%d]", i, x[i], y[i], i);
I found the main problem with this program a few days after posting this and the fix would be removing the ; from the for commands, along with some other minor changes. I thought I might add this comment to let you know and now it is working like a charm. The simplest of mistakes fools even the trained eyes. After I found this, I was shocked that no one picked up on this mistake.