How to sum 2 numbers in C language? [closed] - c

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I am new to programming. I am learning how to sum in C language. Please see below code, What am I missing? Why its giving error?
#include <stdio.h>
int main() {
int a = 1, b = 2, c = 3, d = 0;
d = a + b + c;
printf(d);
return 0;
}

The "f" in "printf" means "formatted". The first parameter you pass to it has to be the format that you want to print with, not what you want to print. In this case, it seems to me that you likely want to:
printf("%d\n", d);
The %d means that printf should interpret the second parameter as a signed integer (which d actually is). The \n adds a newline (and usually flushes the buffer).
You can learn more about printf and its format by googling for it, or reading a man page about it, or its page in a compiler help file.

you are printing a integer value. In order to print a integer value you have follow this way...
printf("%d",d); // for integer it's %d
// so your progam should looks like this
#include <stdio.h>
int main() {
int a = 1, b = 2, c = 3, d = 0;
d = a + b + c;
printf("%d\n", d);
return 0;
}

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why doesn't this program in C language print anything? [closed]

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why doesn't this program print anything?
int main(){
int a = getchar()-'0';
getchar();
int b = getchar()-'0';
int vsota = 0;
vs = (a+b)%10;
putchar(vs);
printf("\n");
}
I put in the numbers 7 and 9 and it was supossed to outprint 6 but it does not.
putchar(vs); writes the character whose code is the value in vs. The value in vs is 6. So it writes the character with code 6. That character is not the digit “6”. You do not see anything because it is a “control character” with no visible appearance. To write the character for the digit whose value is in vs, use putchar('0' + vs);.
Also, fix this:
int vsota = 0;
vs = (a+b)%10;
That would not have compiled, so presumably you made a mistake when entering code into Stack Overflow. Use the same name in both places.
because you try to output non printable character
int main(){
int a = '1'-'0';
int b = '6'-'0';
int vsota = 0;
vsota = (a+b)%10;
putchar(vsota + '0');
printf("\n");
}
https://godbolt.org/z/NAyrNn

How to write (a^n-1) in c programming [closed]

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I want to write (a^n-1) in c programming. I wrote pow(a,n-1) and the output is
wrong. How can I solve this problem? Thanks.
Here is my code:
#include <stdio.h>
#include<math.h>
int main() {
float a, r,n;
float sum = 0;
a = 1.04*pow(a, n-1);
r = 1.02*pow(1.04,-1);
n = 2;
sum = 360000*pow(1.04,n)-50000*(a * (1 - pow((r), n ))) / (1- (r));
printf("\n%.2f", sum);
return 0;
}
The correct output should be 286376 but the program showed 2903773
Every C program executes line by line. So, at the time when compiler came on
a = 1.04*pow(a, n-1);
this line, variable a and n was not assigned with any value, resulted in giving you a garbage value...
So, the problem is, you had not assigned values in variables, and still, you were using them.
You have to first assign values in variables before using them. Otherwise, they will pick any garbage value from memory (Any Random number).
Assign value in a and n and try again.
Edit: As chux's comment suggests, if your program is supposed to give 286376 as output, then value of a should be 1 and value of n should be 2.
So, your correct code would be this:
#include <stdio.h>
#include <math.h>
int main() {
float a=1, r, n=2;
float sum = 0;
a = 1.04*pow(a, n-1) ;
r = 1.02*pow(1.04,-1);
sum = 360000*pow(1.04,n)-50000*(a * (1 - pow((r), n ))) / (1- (r));
printf("\n%.2f", sum);
return 0;
}
pow(a, n-1) translates into exp(log(a) * (n-1)) which isn't precisely the same.. You can try to round the output like this:
round(pow(a, n-1))

Write a function receives two arguments and returns the difference [closed]

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Write a C-program that calls a function minus(). This function receives two arguments and returns the difference (regular subtraction, not absolute). This difference should be printed on screen.
My code is:
int minus(int a,int b)
{
int c = a - b;
return c;
}
int main()
{
int a = 4; int b = 5;
minus(a,b);
printf("%d", minus);
return 0;
}
I have two questions:
1.why a and b in
int minus(int a,int b)
are grey in Visual Studio? "int" is blue but a and b are grey.
2. I got this result but it should be -1.
Could someone help me please
a and b are grey because the editor is automatically coloring the code to help illustrate the program syntax. This may look funny at first, but, over time, your brain will become accustomed to it, and things that are the wrong colors will stand out. This will help you find mistakes in your program—when you make a mistake typing, something that should be a keyword will be colored like a parameter name, and you may notice it is the wrong color and take a closer look at what you typed.
In printf("%d", minus);, the minus is just the function. It is not the value returned by the function. To print the value returned by the function, use printf("%d", minus(a, b));.
You have undefined behavior here printing the function pointer using %d format specifier. (You have used the wrong format specifier that's why the Undefined behavior).
And the most probable way you would like is to printf("%d", minus(a,b));. You wanted to print the result of the subtraction not the function pointer itself.
You are missing an assignment
int minus(int a,int b)
{
int c = a - b;
return c;
}
int main()
{
int a = 4; int b = 5;
int d = minus(a,b);
printf("%d", d);
return 0;
}
you called the function minus() but you did not take the value returned by function in any variable so when you try to print minus then it will return the pointer value of function. so to get the correct answer hold the return value in variable and then print it int c = minus(a,b);
printf("%d", c); or you can call the minus function inside the print function like this printf("%d", minus()); as a beginner i will suggest you to implement the first suggestion it will increase the capacity to use statements in c.

Can any one explain why this section of code is giving run time error? [closed]

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I am not finding the correct output of this program.It giving run time error.
#include <stdio.h>
int main()
{
int c = 5, no = 10;
do {
no /= c;
} while(c--);
printf ("%d\n", no);
return 0;
}
It's division by zero. Since you are using a post-decrement in your loop counter c, it is becoming 0 in the last iteration.
Now that you know the reason for the run time error from the answer by #EugeneSh, here's how you can fix it.
do {
no /= c;
} while(--c); // Use pre-increment instead of post-increment.
In addition the all these answer above I just want to say it's better to check whether a number is zero before division -
#include <stdio.h>
int main()
{
int c = 5, no = 10;
do {
if(c!=0){
no /= c;
}
} while(c--);
printf ("%d\n", no);
return 0;
}
This will prevent these kind of runtime error.
Hope it will helps.
Thanks a lot.

Binary To Decimal [closed]

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The program is suppose to convert binary number to decimal form. Only using scanf() and printf() library functions. Takes in a char array from user ---no prompt outputs decimal form, function must be used with parameter (char binaryString[]) after conversion result must be printed out in main. Program does not work don't think I'm converting the binary form to decimal form correctly in function binaryToDecimal since i cant use pow() I'm lost
#include <stdio.h>
#include <math.h>
int binaryToDecimal(char binaryString[]) {
int c, j = 1, decimalNumber = 0;
for (binaryString[c = 0]; binaryString[c] > binaryString[33];
binaryString[++c]) {
while (binaryString[c] != 0) {
remainder = binaryString[c] % 10;
decimalNumber = decimalNumber + remainder * j;
j = j * 2;
binaryString[c] = binaryString[c] / 10;
}
}
return decimalNumber;
}
int binaryToDecimalMain() {
int arraysize = 33;
char binaryString[arraysize];
scanf("%32s", binaryString);
printf("%d",binaryToDecimal(binaryString []);
return 0;
}
I not give you the algorithm because it's seems that you are learning how to program and it is important to you to learn to discover how to solve the problems that are given to you.But I can give you some hints:
use binaryString only to compare with '0' or '1'. Don't try to make any operations like '%' on it.
iterate on the binaryString character by character (no while inside for [this is only for this case, there some algorithm that is necessary to do something like this])
your logic to convert is on the right track
Also you should call your main function main.

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