I'm running some simple code in my CodeBlocks and I wonder why scanf function cannot work with shorts correctly!
The code below is an example. The code takes from the user three int numbers and then prints them again, that simple — but the values printed don't match the values entered.
#include <stdio.h>
int main()
{
short x, y, z;
printf("Please enter three integers! ");
scanf("%d %d %d", &x, &y, &z);
printf("\n num1 = %d , num2 = %d , num3 = %d ", x, y, z);
return 0;
}
The specifier %d is only used for int variables, but in case of short you must use the %hi specifier instead of %d.
So your code must be :
#include <stdio.h>
int main() {
short x , y , z ;
printf("Please Enter three int Numbers ! ");
scanf("%hi %hi %hi",&x,&y,&z);
printf("\n num1 = %hi , num2 = %hi , num3 = %hi ",x,y,z);
return 0;
}
You can find more information about the C data types and their Specifiers here :
https://en.wikipedia.org/wiki/C_data_types
short != int
you pass the pointer to the (usually 2 byte) data, and scanf expects and writes 4 bytes
change short x , y , z ; to int x , y , z ;
as always scanf is not buggy, but the coder is :)
PS Forgot to add. you can also use h format modifier. There is hh as well if you want to scan char sized variables
Related
I have compiled this code and it works just fine up to value 4 then it starts returning character instead of integer. I am talking about first equation => x= num*2; Here when I enter num value as 5 the output returns a.
#include <stdio.h>
int main(void)
{
int num;
int x; This right here is an integer still it returns a character
char s[10] = "helloworld";
char f[10];
scanf("%d", &num); //
//printf("%d\n", num);
x = num * 2 ;
printf("%x\n", x);
scanf("%c", &f[10]);
if(s[10] = f[10]){
printf("helloworld");
}
}
please tell me if there is a mistake I am a newbie to coding.
As I see you are learning C language, and after reading your explanation, I feel that you want to print the integer value of variable x.
Kindly replace %x with %d in the print statement of variable x,
and you will be successfully able to print the value.
#include <stdio.h>
int main(void)
{
int num;
int x; // This right here is an integer still it returns a character
char s[10] = "helloworld";
char f[10];
scanf("%d", &num);
x = num * 2 ;
printf("%d\n", x); // %d for integer and %x for hexadecimal values
scanf("%c", &f[10]);
if(s[10] = f[10]){
printf("helloworld");
}
return 0;
}
Finally, do read more about format specifiers in scanf and
printf statements.
Hi I am studying C language by myself.
My question is, is there something I missed which need to know when working with scanf() that takes char value?
To practice do...while loop, I wrote some code like below but it did not work as I expected.
#include <stdio.h>
int main()
{
char y_or_n;
int x =1;
int y;
do
{
printf("ENTER A NUMBER\n");
scanf("%d", &y);
printf("THE NUMBER WILL BE ADDED TO x WHICH IS %d\n", x);
x = x+y;
printf("x TURNED INTO %d\n", x);
printf("KEEP DOING THIS?(y/n)\n");
scanf(" %s", &y_or_n);
printf("x is %d\n", x);
}
while(y_or_n =='y');
printf("GOOD BYE\n");
return 0;
}
For the first loop, it worked as I expected.
For example, when I entered 7, x turned into 8. But after scanf() was executed, value of x was changed into 0.
So from second loop, value of x changed temporarily into value of y and changed again into 0.
I guessed that there is something wrong with scanf() function and modified the code slightly: changed type of y_or_n into integer so that scanf() takes integer value.
The modified code is like below
include <stdio.h>
int main()
{
int y_or_n;
int x =1;
int y;
do
{
printf("ENTER A NUMBER\n");
scanf("%d", &y);
printf("THE NUMBER WILL BE ADDED TO x WHICH IS %d\n", x);
x = x+y;
printf("x TURNED INTO %d\n", x);
printf("KEEP DOING THIS?(y/n)\n");
scanf(" %d", &y_or_n);
printf("x is %d\n", x);
}
while(y_or_n ==1);
printf("GOOD BYE\n");
return 0;
}
This time the code worked as I expected.
Value of x was not changed into 0 even after an execution of scanf() and every time I entered a number that number was added to x.
If my question is not clear, please let me know.
Thank you for reading.
%s is for reading null-terminated strings, so passing pointer to one-byte buffer for that is bad.
It seems the variable x is placed just after the variable y_or_n on the memory and writing of terminating null-character by scanf() is setting value of x to 0.
To read one character, use %c instead.
char y_or_n;
/* ... */
scanf(" %c", &y_or_n);
#include <stdio.h>
#include <stdlib.h>
int main() {
int i;
int mult;
int n;
int ans;
ans = mult * i;
printf("Please enter a multiple you want to explore.");
scanf("%d", &mult);
printf("Please enter the number which you would want to multiply this number till.");
scanf("%d", &n);
for(i = 0; i<n; i++) {
printf("%d x %d = %d \n", &mult, &i , &ans);
}
return 0;
}
Hi guys, this is a simple code which is supposed to help the user to list the times table for n times. However, i am receiving undefined behaviour and I am quite stumped as to what is wrong with my implementation of my "for" loop.
I am receiving this as my output.
6356744 x 6356748 = 6356736
for n times in my consoles.
I want to ask
Is anything wrong with the logic of my code? (i assume i do have a problem with my code so please do enlighten me)
Would it be better(or even possible) to use pointers to point to the memory addresses of the mentioned variables when i have to change the value of the variables constantly? If yes, how do i go around doing it?
Thanks!
In printf you must provide integers. You are now giving the addresses of integers. So change
printf("%d x %d = %d \n", &mult, &i , &ans);
to
printf("%d x %d = %d \n", mult, i, ans);
and to make the table, replace ans with just mult*i, so:
printf("%d x %d = %d \n", mult, i, mult*i);
You should also check the return value of scanf to check if it has succeeded reading your input:
do {
printf("Please enter a multiple you want to explore.");
} while (scanf("%d", &mult)!=1);
do {
printf("Please enter the number which you would want to multiply this number till.");
} while (scanf("%d", &n)!=1);
The things you see are the values of the variables memory location.
Change your lines inside for loop as below
ans = mult * i;
printf("%d x %d = %d \n", mult, i, ans);
There are some mistakes in your code .
you are using the & operator in print statement which is used to print the address of the variable.
Initiate the loop with the value '1' instead of '0' & execute the loop till 'i' less than equal to 'n'.
instead of using the ans variable outside the loop , use it inside the loop as it evaluate the multiplication result in each iteration of the loop.
#include <stdio.h>
int main()
{
int i;
int mult;
int n;
int ans;
printf("Please enter a multiple you want to explore.");
scanf("%d", &mult);
printf("Please enter the number which you would want to multiply this number till.");
scanf("%d", &n);
for(i = 1; i<=n; i++) {
ans = mult*i ;
printf("%d x %d = %d \n", mult, i , ans);
}
return 0;
}
I was writing a C Program to find if a number is prime or not. Everytime I run it and enter a number, the value of the input changes. PLease point out the loopholes.
#include<stdio.h>
#include<conio.h>
int main()
{
int x;
int y;
y=getchar();
for(x=2;x<y;++x){
if(y%x != 0 && y!=x)
printf(" THE NUMBER %d is A PRIME \n", y);
else
printf(" \r THE NUMBER %d IS NOT A PRIME", y);
break;
}
}
I use the Code::Blocks IDE with GCC Compiler
As the name implies, getchar() gets a single character from standard input. For example, if you enter 3, the y gets the ASCII code of the character '3', which is obviously not what you want.
Try scanf:
scanf("%d", &y);
getchar returns the ASCII code of a single character. Consequently, your program picks up the ASCII code of the first character of the number you input and checks if it is prime.
Instead, you need to read an integer:
scanf("%d", &y);
The complete program:
#include<stdio.h>
#include<conio.h>
int main()
{
int x;
int y;
scanf("%d", &y);
for(x=2;x<y;++x){
if(y%x != 0 && y!=x)
printf(" THE NUMBER %d is A PRIME \n", y);
else {
printf(" \r THE NUMBER %d IS NOT A PRIME", y);
break;
}
}
}
Note: You can stop when x >= sqrt(y)
Well, you are calling getchar() which is used to input a single character and this is what happens in your case:
getchar() returns a character.
Character is then converted into integer when you store it in variable of type int.
Hence that integer contains the ASCII of input character i.e. 3 will be stored as 51 that is the reason input changes.
What you need to do is to input an integer instead of character. Try this:
scanf("%d", &y);
Hope this helps.
First answers are correct about input for y:
scanf("%d", &y);
Also, please note that you should loop until square root of x, and not more if your want to optimize your algorithm (I won't demonstrate here why, it's a mathematical property).
#include <stdio.h>
#include <math.h>
// ...
int x;
int x_max;
int y;
scanf("%d", &y);
x_max = (int)floor(sqrt(y));
for(x=2;x<=x_max;++x){
// ...
I have an error code I do not understand:
format %d expects type int, but argument 2 has type int *
I do not know the difference between int and int *. I did not know there were different types of int, and cannot find any note of it on webpages about printf and scanf key letters.
The code is as follows:
#include <stdio.h>
#include <math.h>
int main(void)
{
int X = 0, Y = 0, A = 0, D = 0;
printf("This program computes the area of a rectangle ");
printf("and its diagonal length.");
printf("Enter Side 1 dimentions: ");
scanf("%d", &X);
printf("Enter Side 2 dimentions: ");
scanf("%d", &Y);
/* Calc */
A = X * Y;
D = pow(X,2) + pow(Y,2);
D = pow(D, 1 / 2);
/* Output */
printf("Rectangle Area is %d sq. units.", &A);
printf(" Diagonal length is %d.", &D);
return 0;
}
The error references the last two printf's:
printf("Rectangle Area is %d sq. units.", &A);
printf(" Diagonal length is %d.", &D);
Additionally, this program was originally written using floats (declaring X,Y,A, and D as float and using %f). But that gave an even stranger error code:
format %f expects type double, but argument 2 has type float *
I knew that %f is used for doubles and floats, so I could not understand why I had this error. After I got the error code about floats/doubles I tried changing everything to int (as shown in the above code), just to check. But that delivered the error code at the top of this post, which I do not understand either.
I've been using the gcc compiler.
Would someone explain what's being done wrong?
The problem is that you're trying to pass pointers to the printf function. Here's what your code looks like:
printf("Rectangle Area is %d sq. units.", &A);
printf(" Diagonal length is %d.", &D);
A is the int variable, but &A is a pointer to the int variable. What you want is this:
printf("Rectangle Area is %d sq. units.", A);
printf(" Diagonal length is %d.", D);
int* means a pointer to an int object. this is what you get because you use & before the variable name (i.e &A in your code)
You can read this to understand more about pointers and references, but basically if you omit the & before the variable names, it will work fine.
Why passing pointers to printf("...%d...", &D)?
Take a look to pointers explanation:
http://www.codeproject.com/Articles/627/A-Beginner-s-Guide-to-Pointers
And to simplified printf() manual:
http://www.cplusplus.com/reference/cstdio/printf/
int d = 1;
printf("I'm an integer: %d", 42); // OK, prints "...42"
printf("I'm an integer too: %d", d); // OK, prints "...1"
printf("I'm a pointer, I have no idea why you printing me: %p", (void*)&d); // OK, prints "...<address of d>", probably not what you want
printf("I'm compile-time error: %d", &d); // Fail, prints comiper error: 'int' reqired, but &d is 'int*'