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I was trying to explain to my friend something about C coding and he asked me why his code (with "scanf") didn't work.
#include
int main() {
char x=scanf("%c",&x);
printf("%c\n",x);
return 0;
}
and this one yes
#include <stdio.h>
int main()
{
int k;
char x=getchar
printf("%c\n",x);
return 0;
}
When scanf completes, x contains the character that was read. However, that value is immediately overwritten when x is assigned the return value of scanf, which is the number of items successfully matched or EOF in the event of an error.
If you call scanf without assigning the return value to x you should get the expected result.
For example, this should work.
char x;
scanf("%c",&x);
Related
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Hi can anyone help me to correct this code, the result should be
/c=4.000000/
/d=4.0000 /
I know by putting the logic in single printf() i will get my result but i am not understanding that how to use two printf() and the varibles will be given by the second printf().
Here is my code:-
#include<stdio.h>
int main()
{
int a=19,b=4;
float c,d;
c=a/b;
d=a%b;
printf("/c=%12f/\nd=%");
printf("-12.4f/",c,d);
putchar(10);
return 0;
}
If I change my code to this, I will get the result,
#include<stdio.h>
int main()
{
int a=19,b=4;
float c,d;
c=a/b;
d=a%b;
/*
printf("/c=%12f/\nd=%");
printf("-12.4f/",c,d);
*/
printf("/c=%12f/\n/d=%-12.4f/",c,d);
putchar(10);
return 0;
}
but I want to use two printf() statements.
Thanks in advance.
You can not do this:
printf("/c=%12f/\nd=%");
printf("-12.4f/",c,d);
because you are lying to both printfs, in the first one you don't use the specifiers and in the second one you use specifiers that are not expected.
You can do this:
printf("/c=%12f/\nd=%"
"-12.4f/",c,d);
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I am new to C. I was just interested in making a program which takes two integer input from the user and add it but the condition is that i have to use only one variable. I came up with this code:
#include <stdio.h>
int main()
{
int *a;
scanf("%d%d",a,(a+1));
printf("\nSum=%d",*a+*(a+1));
return 0;
}
scanf() function takes an valid address as an argument and i am passing the value in a(which is not initialised yet). So, how this code worked in Turbo C++?
You are trying to access an area that is not within the scope of the program. Luckily TCC gave it, but I believe if you go on experimenting, results will be undefined.
You can do something like this to solve your problem of adding using 1 variable.
int main()
{
int a;
scanf("%d",&a); //scan the first number
getchar();
a += getchar()-'0'; // get the second number (restricted to 0-9)
printf("%d",a);
return 0;
}
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Before I start please keep in mind that I am still learning and need help with some stuff that may be easy to you but not to me. So here we go. I am having trouble using scanf in a custom function. It will not let me type anything. It just keeps running forever unless I stop it. How can I get scanf to work here:
#include <stdio.h>
#include <stdlib.h>
void function ();
int main (void)
{
char sel;
function ();
return 0;
}
void function ()
{
scanf("%c",&sel);
}
There is no problem with scanf. The problem is with your loop. l = 1 and l++ will always keep l >= 0 and hence it will keep on taking input infinite times.
Also, it will keep on overriding the value of var.a with every input
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I've found this function on the internet and I find it to be very useful
but I am new to programming, and could someone please explain briefly what does it exactly do
#include <stdio.h>
int diffcount(char* s)
{
unsigned char seen[127];
int cnt=0,i;
for(i=0;i<127;i++)
seen[i]=0;
for(i=0;s[i];i++)
{
if(!seen[(int)s[i]])
{
cnt++;
seen[(int)s[i]]=1;
}
}return cnt;
}
int main(void) {
char string[20];
scanf("%s",string);
printf("Razlicitih znakova: %d\n", diffcount(string));
return 0;
}
First of all we init an empty array of zeros int seen[127];
"seen" array is used to find out whether char with code i has been met in the array s : if seen[i]==1 than (char)i was in the string s.
After that we make a loop through char* s and check if char s[i] has already been met by looking at the value of seen[s[i]]; and if it is false we put seen[s[i]]=true (because we met it) and increase our counter.
The result of the function is the value of variable cnt
This may also help:
each char has it's code between zero and 127. For example, (int)'a' = 97.
bool in the C is just the same as int, that's why we sometimes use 0 and 1 instead of true and false
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Write a C program that reads lines containing floating point values of type double one per line from the standard input ( e.g. using scanf ), converts those values to integers and then prints those values as right justified integers in a 20-character wide field one per line on the standard output.
#include <stdio.h>
My biggest problem is I don't know where to start. Any tips and help would be appreciated.
The concept is to TYPECAST the float to an integer.
The loop here is for multiple values if you want.
This is the program. I hope this helps; it runs as you want.
#include <stdio.h>
int main()
{
float n;
int t;
//loop here
scanf("%f", &n);
t = (int)n;
printf("%20d", t);
// end loop here
return 0;
}