I'm new to C language. Here's the code I used to get input for a and b and print them. But I did not get the values I entered via the terminal can you explain why I got different values for a and b?
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
int a,b;
scanf("%d, %d", &a, &b);
printf("%d, %d", a, b);
return 0;
}
I entered 5 9 as inputs, but I got 5 and 16 as outputs.
You need to check the value of scanf() otherwise some or all the values you attempt to use are undefined. In your case the input you provided did not match the scanf() format string. I changed the format string to match the input you provided, removed headers that are not used and added a newline to your printf() statement:
#include <stdio.h>
int main(void) {
int a, b;
if(scanf("%d%d", &a, &b) != 2) {
printf("scanf failed\n");
return 1;
}
printf("%d, %d\n", a, b);
}
You don't have to use comma (,) to separate 2 inputs in scanf() function. This function automatically separates your inputs by number of occurrences of format specifiers i.e. %d in your case. However you have to separate variable addresses using comma.
Just use it like scanf("%d %d", &a, &b).
Related
I recently started learning C and must create program that scanf two integer values from standard input separated by a space then printf the sum of these two integers. Must be able to accept negative values. I'm using repl.it to write code 1st then pasting in .c to compile.
Attempt:
#include <stdio.h>
int main(void) {
int j = 0;
printf("Enter 2 integers separated by space and press enter:\n");
while (scanf("%d", &j) == 1) {
printf("Here is the sum:"%d", j);
}
return 0;
}
Except this prints
Here is the sum: 1Here is the sum: 2
The output is wrong so what mistake did I make? Whats the correct method to get expected values?
(eg 1+2=3)
Your code is incorrect. It does not even compile due to "Here is the sum:"%d" ill formatted string. The program will repetitively scan an integer from standard input and print it as it was result of a sum.
Parsing two integers separated by a space can be easily done with scanf() function by using "%d %d" pattern. The function scanf() returns a number of successfully parsed arguments thus value of 2 is expected on correct input. Finally, add both number and print the result.
#include <stdio.h>
int main(void) {
int i, j;
if (scanf("%d %d", &i, &j) == 2) {
printf("Here is the sum: %d\n", i + j);
return 0;
}
I executed following code in ubuntu with gcc compiler.
As a=0, the second printf() prints some garbage value.
What kind of behavior is it by printf()?
#include <stdio.h>
#include <stdlib.h>
int main(void) {
int a = 0;
printf("\nThe Value of %ns : %d\n", &a, a);
printf("\n%d\n", a);
printf("\n\n");
return 0;
}
If you read e.g. this printf reference you will see that the %n format specifier will:
returns the number of characters written so far by this call to the function.
So it will overwrite the contents of a with the number of characters it has written so far, which should be 14 if I count correctly.
The output of the first printf is 0
Because the second argument of printf passes a by the value and it gets printed afterwards.
The output of the second printf is 14
Because at this time the value of a was replaced by the number of characters that was printed before %n by the first printf
As a=0, the second printf() prints some garbage value.
You can't believe that statement until you verify it!
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
int main(void) {
int a = 0;
printf("\nThe Value of %ns : %d\n", &a, a);
assert(a == 0);
printf("\n%d\n", a);
printf("\n\n");
}
Any time you want to say "I'm sure that (some invariant) is true at place x", use assert to state that fact and let it be checked at runtime.
Output:
output.s: ./example.c:8: main: Assertion `a == 0' failed.
Only my first line of input request user to key in the value. Input B not request user to key in and shows wrong total.
#include <stdio.h>
#include <stdlib.h>
//BASIC CALCULATION INPUT 2 INTEGER ONE BY ONE
int main(int argc, char *argv[])
{
int a,b,c;
//REQUEST ONE INPUT
printf("Integer A: \n");
scanf("%a",&a);
//REQUEST ONE INPUT
printf("Integer B: \n");
scanf("%b",&b);
c=a+b;
//DISPLAY AMOUNT INTEGER
printf("Total: &c",c);
system("PAUSE");
return 0;
}
Your both scanf statements are wrong!
It should be
scanf ("%d",&a);
scanf ("%d",&b);
For taking user input a and b of type integer use %d.
&a is the reference (address) of identifer a which holds the value of a.
And also user output printf statement for integer c should be
printf("%d",c);
Add a space in the scanf to discard all whitespace before matching an integer. Such as a
scanf(" %b",&b);
^^^
space in the scanf
There is no format specifier like %b that's why Input B doesn't request user to key in.I think this this will help you.
#include <stdio.h>
#include <stdlib.h>
//BASIC CALCULATION INPUT 2 INTEGER ONE BY ONE
int main(int argc, char *argv[])
{
int a,b,c;
//REQUEST ONE INPUT
printf("Integer A: \n");
scanf("%d",&a);
//REQUEST ONE INPUT
printf("Integer B: \n");
scanf("%d",&b);
c=a+b;
//DISPLAY AMOUNT INTEGER
printf("Total: &c",c);
system("PAUSE");
return 0;
}
Both of your scanf statements use wrong format specifiers. Thus undefined behaviour.
scanf("%a",&a);
and
scanf("%b",&b);
a expects a float*as its argument, but you are passingint*. There's no format specifierb` in standard C either.
Use %d to scan int's.
Better yet, avoid scanf altogether and use fgets and parse the line instead.
Another problem is your printf statement:
printf("Total: &c",c);
is wrong too. It should be:
printf("Total: %c",c);
to print the value of c.
#include <stdio.h>
int main ()
{
double a=0;
char b=0;
scanf ("%d%c",&a,&b);
printf ("%d,%c", a, b);
return 0;
}
This is my code for a quick test program I wrote to play around with the scanf function in C. I am trying to have the user input something like 78X + 5 = 19 (then hit enter) and then parse that into variables a, b, and c where in this case a=78, b=5, c=19. In the sample code, when I type in 78X, c doesn't store a value to b and only prints "78, " and then terminates. Why won't it store a value to b?
If your input is 75x then below is the code which reads the value and stores it in a(75) and b(x) respectively
#include <stdio.h>
int main ()
{
int a=0;
char b=0;
scanf ("%d%c",&a,&b);
printf ("%d%c", a, b);
return 0;
}
The , in your format string is significant. The string %d,%c would match the input 78,x but it would not match 78x .
Also you need to use %f to scan and print a double. Using %d causes undefined behaviour (which may manifest itself as b seeming to not appear). Either change to %f, or change your double to an int.
I'm trying to write a short program were:
#include <stdio.h>
void main()
{
char=a,b,c;
printf("please place 3 numbers:\n");
scanf("%c%c%c", &a,&b,&c);
}
The exercise I'm trying to solve is how to change the char to int so if I write in a the number 3, I will get the number 3 Printed.
at this point I'm only getting the value.
I would appreciate any help.
The answer depends somewhat on what you can assume about the character set. If it's something like ASCII (or really, any character set that includes the digits in sequential order), you just need to offset the character value by the value of the character 0:
int aValue = a - '0';
I'm sure that C# provides better ways to do what you're trying to do, though. For example, see this question for some examples of converting strings to integer values.
First of all your syntax need some checking
You should know that you declare a variable this way (a char in this example):
char a;
If you want to declare multiple variables of the same type in a row you do :
char a, b, c;
If you want to assign a value to a declared variable :
a = '3';
Now to print a char using printf (man printf is a must read, more infos are in coreutils) :
printf("%c", a);
If you want to get the char from the command line, I recommand you to use getchar() (man getchar) instead of scanf because if suits better what you are trying to achieve and doesn't require you to use a syntax in scanf that I am sure you don't fully understand yet.
Your question is incredibly light on details, so here are several options:
#include <stdio.h>
int main()
{
char a,b,c;
printf("please place 3 numbers:\n");
scanf("%c%c%c", &a,&b,&c);
printf("Printing ints (auto-promotion): %d %d %d\n", a, b, c);
printf("Printing ints (explicit-promotion): %d %d %d\n", (int)a, (int)b, (int)c);
printf("Printing digits: %d %d %d\n", a-0x30, b-0x30, c-0x30);
return 0;
}
If the input is 123,
I expect the output to be:
Printing ints (auto-promotion): 49 50 51
Printing ints (explicit-promotion): 49 50 51
Printing digits: 1 2 3
Some things I fixed along the way.
main should return an int, not be void.
char=a,b,c; is a syntax error. You meant char a,b,c;
added a return 0; at the end of main.
You question is not quite understandable. Still I'll try to help. I think that what you want is to store an integer value in the char variable. You can do so by using the following code:
#include<stdio.h>
void main()
{
char a,b,c;
printf("Enter three numbers:\n");
scanf(" %c %c %c",&a,&b,&c); //notice the spaces between %c
}
Or if you want to enter a character and print its ASCII value, you can use the following code:
#include<stdio.h>
#include<conio.h>
void main()
{
char a,b,c;
printf("Enter three characters:\n");
scanf(" %c %c %c",&a,&b,&c);
printf("Entered values: %d %d %d",a,b,c);
getch();
}