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For an assignment I need to convert a 16-bit decimal number to a binary number. So for example the number 9 should print 0000000000001001. My professor started us with this code:
void printBinary(short n)
{
}
int main(int argc, char **argv)
{
short n;
printf("Enter number: ");
scanf("%hd", &n);
printBinary(n);
}
I am very confused as to where to go from here. I really would appreciate anyone helping me understand what to do, as I am very new to coding. Thanks in advance.
Try this:
int printBinary(int integer)
{
int remainder, counter = 0, my_binary = 0;
while (integer > 0)
{
remainder = integer % 2;
my_binary += remainder * pow(10, counter);
integer /= 2;
counter++;
}
printf ("Binary is %d\n", my_integer");
}
This has been taken from here.
NOTE: I have the assumption that you want to convert only positive decimals.
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I want to use atof to convert my string to a double，the answer is correct ,but not very accurate
ATTENTION: because of some other reasons, fscanf is not permitted
my code is :
#include <stdio.h>
#include <stdlib.h>
#define MAXCN 50
int main(void)
{ FILE* lstm_txt = NULL;
char lstm_weight[MAXCN] = {0};
int lstm = 0;
int i = 0;
float lstm_val;
if ((lstm_txt = fopen("test1.txt", "r"))== NULL){
fprintf(stderr,"error:file open failed 'test1.txt'.\n");
return 1;
}
while ((i + 1 < MAXCN) && ((lstm = fgetc(lstm_txt)) != ' ' ) && (lstm != EOF)){
lstm_weight[i++] = lstm;
}
//lstm_weight[i] = 0;
printf("\n lstm_weight: %s\n\n", lstm_weight);
lstm_val = atof(lstm_weight);
printf("\n convert \"lstm_weight\" to lstm_val is : %f\n\n", lstm_val);
return 0;
}
my file : lstm_txt is ：
4.217959344387054443e-01 -2.566376626491546631e-01 2.173236161470413208e-01 4.217959344387054443e-01
code hasn't bug, and the result is :
lstm_weight: 4.217959344387054443e-01
convert "lstm_weight" to lstm_val is : 0.421796
but I want Istm_val is 0.4217959344387054443 ,how can I do that?
you could try something like sprintf()
here's an example:
#include <stdio.h>
int main() {
char str[50];
double n = 0.3984092590879;
sprintf(str, "%lf", n);
printf(str);
return 0;
}
prints out:
0.3984092590879
Printing %.17f you can have a precision up to 0.42179593443870544
printf("\n convert \"lstm_weight\" to lstm_val is : %.17f\n\n", lstm_val);
A double typically has about 15 digits of decimal precision. You will not get more accuracy than that if you store the value in a double. You are getting less because you didn't tell printf how many digits of precision to use for output, so you got the default.
Use something like %0.15f instead of %f.
convert "lstm_weight" to lstm_val is : 0.421795934438705
With %0.20f, I get:
convert "lstm_weight" to lstm_val is : 0.42179593443870544434
That's the best you'll do with a double.
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I will directly get to the point.
int a,b,c, Number;
a=0;
b=1;
c=2;
Number = (a*100)+(b*10)+(c*1);
printf("%d", Number);
Result is 12.
I also have another variable that which is equal to exactly 12.
int number2 = 12;
Now I have to check the difference between them. How do I do that?
Just use a if statement like below:-
int diff = 0;
if ( Number > number2 )
{
diff = Number - number2;
printf("Number is greater than number2 by %d\n", diff);
}
else if( number2 > Number )
{
diff = number2 - Number;
printf("number2 is greater than Number by %d\n", diff);
}
else
printf("Numbers are equal \n");
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Calculating e^x but I'm really new to programming. I've made like two programs before. I've fixed any errors the shell points out yet get zero once it is run.
I have this :
#include <stdio.h>
int main (void) {
float answer = 1.0, x, next_term, n;
int power;
printf("Enter x: ");
scanf("%f", &x);
printf("Enter the number of terms: ");
scanf("%f", &n);
next_term = x;
for (power = 1; power <= n ; power = power + 1) {
answer = answer * next_term;
next_term = next_term * x / (power + 1);
}
printf("exp(%f) = %f\n", x, answer);
return 0;
}
Output:
exp(8.000000) = 0.000000
If you're calculating an infinite series, you want to add the terms, not multiply them... i.e. answer=answer+next_term;
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I need to create a program that gives the sum of the all the numbers between the constants of a and b given by the user. b needs to be greater than a.
#include <stdio.h>
void main()
{
int index, begno, endno, sum = 0;
printf("Program for sum of all numbers in the given range\n");
printf("Enter Beg. No.: ");
scanf("%d", &begno);
printf("Enter End. No.: ");
scanf("%d", &endno);
index = begno;
for(; index <= endno; index ++)
sum = sum + index;
printf("The sum of even numbers between %d and %d is: %d", begno, endno, sum);
}
The code given looks OK, but if you want the sum without including the last number, as is generally the case you should change the for loop like this
for(; index < endno; index ++)
I would start by implementing a loop to compute:
$$\sum_{n=a}^{b}
n$$
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I'm a beginner and I am having a really hard time while doing this program.
The question is:
(1/1!)+(2/2!)+(3/3!)+(4/4!)- - - -n
So here are the n number of terms(in which a number is divided by its factorial) and I have to display the output of the sum of any number of terms which are given in scanf function.
Only one thing I know is that this program can be done by using "Nested for" loop but I haven't perfect grip yet on C language. So you guys have to have help me out in this. :)
#include <stdio.h>
#include <conio.h>
void main(void){
int s,a,b,n,fact=1;
//clrscr();
printf("Enter number of terms=");
scanf("%d",&n);
for(a=1;a<=n;a++) {
fact=fact*a;
b=(a/fact);
printf("Sum=%d",s);
}
getche();
}
P.S It's must for me to do it with "Nested for" loop.
No you do not need any Nested for loops to solve your problem. Here's a procedure you may follow:
function factorial
Input: numbers L.
Output: factorial of L.
function sum
Input: n.
Output: sum.
sum = 0;
for i = 1 to n, do
sum ← sum + (i / factorial(i))
return sum
#include <stdio.h>
int main(void) {
// your code goes here
int n;
float sum = 0,d,fact =1,j,i;
printf("Enter the number:");
scanf("%d",&n);
for(i=1;i<=n;i++){
fact = 1;
for (j = 1; j <= i; j++){
fact = fact * j;
}
d = (float) i / (float) fact ;
sum = sum + d;
}
printf("sum = %f", sum);
return 0;
}
Its working ..you can check over here :-https://ideone.com/JVXQVX