I'm using the following code to get the output up to 5 decimal characters of any number input by user when divided by 1, I have to typecast it with (float).
Can any one tell me how this can be done without typecasting or using float constant?
int main() {
int n;
scanf("%d",&n);
printf("%.5 ", 1/(float)n);
return 0;
}
You can use this piece of code that uses only integers:
printf(n==1?"1.00000":"0.%05d ", 100000/n);
Taking your question strictly literally, you could do:
int main() {
float n;
scanf("%f",&n);
printf("%.5f", 1/n);
return 0;
}
In this code, there is no float literal and no (float) cast.
Technically speaking, this should do it:
printf("%.5f ", 1/pow(n,1));
Related
Why is this code not running after printing of array if I take value of n>=9?
#include <stdio.h>
#include <math.h>
float mean_function(float array[],int n);
int main() {
int i,n;
float array[n],mean,sum=0,s2,summation,deno,C[i],elements;
printf("Enter No of Elements\n");
scanf("%d",&n);
printf("Enter Elements\n");
for(i=0;i<n;i++){
scanf("%f",&array[i]);
printf("%f",array[i]);
}
printf("sample variance(s2) : (sum((x-mean)*(x-mean)))/(n-1) /n");
printf("population variance(sigma2) : (sum((x-u)*(x-u))/n");
mean_function(array,n);
for(i=0;i<n;i++) {
deno=((array[i]-mean)*(array[i]-mean));
C[i]=deno;
summation=summation+C[i];
}
s2=((summation)/(n-1));
printf("s2=%f \n",s2);
}
float mean_function(float array[],int n) {
int i;
float sum=0,mean;
for(i=0;i<n;i++){ sum=sum+array[i]; }
mean=(sum/n);
return mean;
}
Why is this code not running after printing of array if I take value
of n>=9?
Some thoughts about your code (and about building programs in steps):
Arrays in C don't change in size once defined. VLAs are out for a variety of reasons. malloc() is in.
Use double, unless there is a specific reason to use floats.
Define and initialize one variable per line. Uninit vars can only result in an error as mentioned by #Jens.
Function declarations at the top (which you have done)
During development, there is no need to complicate things with a scanf (at least initially). It only adds an unwarranted layer of complexity. If you are testing statistical functions (mean, variance), put numbers in a pre-defined static array and verify functionality first.
C[i] as been declared with uninitialized i.
For this initial phase of building this program, I include a basic program.
I am not a fan of zero space between tokens (but ignore that)
Consider calling your array something other than 'array'.
Calculating the size of the samples array allows you to change the number of elements without changing anything else in code; which adds another layer of complexity to an already difficult phase.
#include <stdio.h>
#include <math.h>
double sample_mean(double* p, int n);
int main()
{
double samples[] = {1.1, 2.2, 3.3, 4.4, 5.5, 6.6, 16.5, 2.3};
double mean = 0.0;
int size_samples = sizeof samples/sizeof(double);
printf("size_samples = %d\n", size_samples);
mean = sample_mean(samples, size_samples);
printf("Mean = %.2lf", mean);
}
// -------------------------------
double sample_mean(double* p, int n)
{
double mean = 0.0;
double total = 0.0;
for(int i = 0; i < n; i++)
total += *p++;
mean = total/n;
return mean;
}
Once this functionality is present (saved), you can start working on other stat functions. This way you can work step by step to get closer to the desired outcome.
Next up you can define sample_variance(double* p, int n) and work on that knowing that additional(new errors) are not coming from your code written so far.
Output:
size_samples = 8
Mean = 5.24
I hope it helps.
The code is likely not running because array[n] is declared with an uninitialized n. At the time you read n with scanf(), the array does not automatically "grow into the right size". You should either declare array big enough, or if you really want it to be user-defined, use malloc to allocate it (read the comp.lang.c FAQ) and all Stackoverflow questions tagged array...)
In addition, the scanf at some point fails. Note that when you enter numbers, you also have the "Enter" as a newline ('\n') in the input stream. You never read the newline so the next scanf fails.
This becomes obvious when you actually check the return value from scanf with code like this:
if (scanf("%f", &array[i]) == 1) {
/* successfully converted 1 item */
}
else {
/* scanf failed */
}
Usually what you want is to skip whitespace in the input. You do this by placing a space in the scanf format. Note that a single space tells scanf to skip any amount of white-space.
if (scanf(" %f", &array[i]) == 1) {
I wrote code for getting binary form of an integer. It works well for inputs like 1 or 10. However, it is failing for inputs like 256. (It gives 0000000 s output and misses the one).
#include <stdio.h>
#include <math.h>
int number_of_binary_digits_required(int n){
return ceil(log(n))+1;
}
void print_array(int * a, int n){
int i = 0;
for (;i<n;i++){
printf("%d\t", a[i]);
}
}
int main(){
int num = 256;
int binary[100];
int n = number_of_binary_digits_required(num);
int bin_digits = n-1;
while (num){
int temp = num%2;
num = num / 2;
binary[bin_digits] = temp;
//printf("%d\n", bin_digits);
bin_digits--;
}
print_array(binary, n);
//printf("%d", number_of_binary_digits_required(num));
//for(bin_digits = 0;bin_digits < number_of_binary_digits_required(num);bin_digits++)
//printf("%d",binary[bin_digits]);
}
Why is the issue coming and how to resolve it?
Thanks you!
C's log function gives result with a base of e, not 2. This is why some numbers give unexpected result in your program since you calculate using that. There is a function log2 which is what you need i think.
Your use of a logarithmic function to compute the number of digits in conjunction with ceil will suffer due to floating point undershoot.
A more reliable way of calculating the number of binary digits is to divide by two repeatedly until zero is attained.
The first mistake is to use log(n), which calculates log of n base e.
Instead use log2(n)
Hope it helps. :-)
Please give me some feedback on how to make my code better or more efficient. It should convert a decimal integer to binary.
#include <stdio.h>
binarydigits(int div, int dis)
{
int numit;
numit=0;
do
{
++numit;
div /= dis;
}
while (div!=1);
++numit;
return numit;
}
main()
{
int x, nb, i;
printf("\n Input an decimal integer number to be converted: ");
scanf("%d", &x);
fflush(stdin);
if (x==0 || x==1)
{
printf("\n\n %d in binary : %d", x, x);
}
else
{
printf("\n\n %d in binary : ", x);
nb = binarydigits(x, 2);
// the function 'binarydigits' returns how many binary digits are needed to represent 'x'
int remind[nb];
// an array of 'nb' elements is declared. Each element of this array will hold a binary digit
for(i=(nb-1) ; i>=0 ; --i, x/=2)
{
remind[i] = x%2;
}
//this 'for' structure saves the remainder of 'x/2' (x%2) in an element of the 'remind[nb]' array
for (i=nb ; i>0 ; --i)
{
printf("%d", remind[nb-i]);
}
//this 'for' structure prints the elements of the 'remind[nb]' array in increasing order
}
getch();
return 0;
}
Any tips on how to make this better would be nice.
Firstly, binarydigits should have a return type int. This is because you return an integer variable numit at the end of this function. Change your function header to:
int binarydigits(int div, int dis)
Secondly, the main() function needs to have a return type int by definition in C, and C++ for that matter. Without it, your compiler will produce a warning, something similar to:
main.c:18:1: warning: return type defaults to 'int' [-Wimplicit-int]
main()
^~~~
Here is a snippet from the the C11 standard (ISO/IEC 9899:2011) on the definition of the main() function:
The function called at program startup is named main. The implementation declares no prototype for this function. It shall be defined with a return type of int and with no parameters: - Return Type of main()
int main(void) { /* ... */ }
Thirdly, you should remove fflush(stdin) because using the fflush() for stdint is undefined behavior as it is not a part of standard C. From C11 7.21.5.2, fflush works only with output/update stream, not input stream:
If stream points to an output stream or an update stream in which the most recent operation was not input, the fflush function causes any unwritten data for that stream to be delivered to the host environment to be written to the file; otherwise, the behavior is undefined. - fflush(stdin)
How to make my code better or more efficient?
My advice to you is to stop trying to learn C by trial-and-error method. You should obtain a good book and study it first. It is impossible to create a fast and efficient C program without mastering pointers, bitwise operators and memory manipulations.
Simply, to make your code fast, you should completelly delete your code (I am not going to list all of your bad-practice things here) and start understanding my example:
int main(void){
char *s = (char*)malloc(33);
char *t = s;
int a;
s += 33;
*s = 0;
printf("Enter a number: ");
scanf("%d", &a);
printf("That number in binary: ");
while(a){
*(--s) = a & 1 ? '1' : '0';
a >>= 1;
}
printf("%s\n", s);
free(t);
return 0;
}
Explanation: we have pointer (if you don't know pointers, well you should probably first learn them) s which points to the end of a string. While number from input (number a) is nonzero, we put its last binary digit in the string, decrease pointer and divide a integrally by 2 (this is a >>= 1 instruction). When a is 0, just print the string.
Below is my code
main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c;
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}
The output was
2 3 4 6 5
How the above code is executed?
it's executed exactly they way you think it's executed: you print the array elements in your for loop, however, you made the array elements as int, therefore, when printing 2.8 which is double, the compiler ignores whatever after the point, means, it sees it as 2 rather than 2.8
When you write the statement :
int c[ ]={2.8,3.4,4,6.7,5};
the decimal values are automatically converted to integers.So the array stores
2,3,4,6,5
So your output becomes as it is.
To work with decimals use float type variables.
float c[ ]={2.8,3.4,4,6.7,5};
The int c[] is integer.You can try it.
float c[] = { 2.8, 3.4, 4, 6.7, 5 };
float *p = c;
for (int j = 0; j<5; j++){
printf(" %f ", *p);
p = p+1;
}
if you use 'float j' instead of 'int j' and '%f' instead of '%d' you would get what you want.
Here due to the using of int j and %d, only the part before the decimal point of each number in the array is printed.
I have this problem with this homework I'm supposed to do.
[ It says Create a program that's able to calculate and show the sum of S]
Like S=1+1/4+1/8+1/16 ... till 1/ [2 pow n]
So I worked on it and came up with this code
#include <stdio.h>
void main()
{
int n,i;
float p,s;
printf("Enter the maximum power n :");
scanf("%d",&n);
s=0;
p=0;
for (i=0;i<n;i++)
{
p+=1/pow(2, i);
s+=p;
printf("s = %f\n",s);
}
printf("The sum of this equation is :%f",&s);
}
But when I execute it is always like S=0.
What am I doing wrong?
You are printing an address ('&s) with%f` variable. Using a wrong specifier invokes undefined behavior. You may get anything.
Also, No need of variable s. Remove the line
s+=p;
It should be like:
#include <stdio.h>
int main(void)
{
int n,i;
float p;
printf("Enter the maximum power n :");
scanf("%d",&n);
p=0;
for (i=0;i<n;i++)
{
p+=1/pow(2, i);
printf("p = %f\n",p);
}
printf("The sum of this equation is :%f",p);
}
You need to include <math.h> to get the proper prototype of pow().
You might need to link to the math library too gcc main.c -Wall -lm
#include <math.h>
....
for (i=0;i<n;i++)
{
p=1/pow(2, i);
s+=p;
printf("s = %f\n",s);
}
printf("The sum of this equation is :%f",s);
Your program has multiple problems. Enabling compiler warnings should tell you about some of them.
You should include the C header which contains the declaration of the pow function.
You add each addend twice.
In your second printf statement, you pass a float. But the %f format specifier expects a double argument. In your third printf statement, you pass a pointer to a float.
Another cosmetic problem is that your main function should return an int.
just a guess, may you replace the line
p+=1/pow(2, i);
with
p+=1.0f/(float)pow(2, i);
and
printf("The sum of this equation is :%f",&s);
with
printf("The sum of this equation is :%f",s);
Typo may be.. but you will have say %f and s (not &s)
printf("The sum of this equation is :%f",s);
On side note:
Once you include <math.h> you will get compiler warning for using correct pow(..) prototype. Below code would be relevant.
p+=1.0f/(float)pow(2.0f, i);