Take for example
int i=10,j;
float b=3.14,c;
char str[30];
sprintf(str,"%d%f",i,b);
sscanf(str,"%d%f",&j,&c);
printf("%d ----- %f\n",j,c);
OUTPUT :- 103 ----- 0.1400000
As you can see, initially i=10 and b=3.14.
I want that j=10 and c=3.14 by using sprint() and sscanf().
The problem I am facing is that the compiler assigns j=103 and c=0.140000.
Is there any way to get rid of this problem in sscanf()?
Add one space to sprintf. Change:
sprintf(str,"%d%f",i,b)
to
sprintf(str,"%d %f",i,b)
Aside: It would be also safer to use snprintf here:
snprintf(str, sizeof str, "%d %f", i, b)
The best way will be to separate the numbers, using a different sign, but if you know that the first int is 2 chars long you can specify it:
sscanf(str,"%2d%f",&j,&c);
// ^^
You are missing a space
Change
sprintf(str,"%d%f",i,b);
to
sprintf(str,"%d %f",i,b);
#include <stdio.h>
int main()
{
int i=10,j;
float b=3.14,c;
char str[30];
sprintf(str,"%d %f",i,b);
sscanf(str,"%d%f",&j,&c);
printf("%d ----- %f\n",j,c);
return 0;
}
output
~ > ./a.out
10 ----- 3.140000
The %d conversion specifier in sscanf will match any number of contiguous numeric characters in the buffer pointed to by str till it encounters a non-numeric character which it can't match. This will cause the integral part of the float to be read as part of the int value. Therefore you must have a way to separate where you integer ends and float starts in the string str. You can put any non-numeric character as a sentinel to separate int value from the float value.
int i = 10, j;
float b = 3.14, c;
char str[30];
// a # to separate the two values, can be any non-numeric char so that it
// is not mistaken for a digit in the int or float value
sprintf(str,"%d#%f", i, b);
// match the separator character to read int and then float
sscanf(str, "%d#%f", &j, &c);
Related
When you declare two variables char a,b; and then you use first 'a' and then 'b',it prints only b, but if you declare it 'b' then 'a', it has no problem printing both in ASCII,the point of the program is to read 121 and 120 and to print yx. the problem - https://prnt.sc/pr5nww
and if you swap them -https://prnt.sc/pr5mt5
#include <stdio.h>
#include <stdlib.h>
int main(){
char a,b;
scanf("%d",&a);
scanf("%d",&b);
printf("%c",a);
printf("%c",b);
}
This is kind of a confusing situation. When it comes to mixing char and int values (as you might do when investigating the numeric values of characters in a character set), it turns out the rules for scanf and printf are almost completely different.
First let's look at the scanf lines:
char a,b;
scanf("%d",&a);
scanf("%d",&b);
This is, in a word, wrong. The %d format in scanf is for scanning int values only. You cannot use %d to input a value of type char. If you want to input a character, the format for that is %c (although it'll input it as a character, not a number).
So you'd need to change this to
char a,b;
scanf("%c",&a);
scanf("%c",&b);
Now you can type characters like A and $ and 3 and have them read into your char variables a and b. (Actually, you're going to have additional problems if you hit the Return key between typing the characters for a and b, but that's a different story.)
When it comes to printing the characters out, you have a little more freedom. Your lines
printf("%c",a);
printf("%c",b);
are fine. And if you wanted to see the integer character-set values associated with the characters, you could have typed
printf("%d",a);
printf("%d",b);
and that would have worked, too. This is because when you call printf (and other functions ike it), there are some automatic conversions that take place: types char and short int are automatically promoted to (passed as) int, and type float is promoted to double. But these automatic conversions happen only for values of those types (as when calling printf). There a=is no such conversion when you're passing pointers to these types, as when calling scanf.
What if you wanted to read numbers, not characters? That is, what if you wanted to input the number 65 and see it get printed as capital A? There are several possible ways to do that.
The first way would be to continue to use %d in your scanf call, but change the type of your variables to int:
int a,b;
scanf("%d",&a);
scanf("%d",&b);
Now you can print a and b out using either %c or %d, and it'll work fine.
You could also use a temporary int variable, before reassigning to char, like this:
char a,b;
int tmp
scanf("%d",&tmp);
a = tmp;
scanf("%d",&tmp);
b = tmp;
The final, lesser-known and somewhat more obscure way, is to use the h modifier. If you say
char a,b;
scanf("%hhd",&a);
scanf("%hhd",&b);
now you're telling scanf, "I want to read decimal digits, but the target variable is a char, not an int."
And, again, you can print a and b out using either %c or %d, and it'll work fine.
the point of the program is to read 121 and 120 and to print yx
Do
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
char a, b;
/* Scan into the half of the half of an int (the leading blank
makes scanf() eat whitespaces): */
scanf(" %hhd", &a);
scanf(" %hhd", &b);
/* Print the half of the half of an int: */
printf("%hhd", a);
printf("%hhd", b);
}
To print the characters literally do the printing part like this:
...
printf("%c", a);
printf("%c", b);
}
This is my code
#include <stdio.h>
int main() {
float percentage;
int sp;
int bp;
percentage = (sp-bp)/bp*100;
scanf("%d %d", &sp, &bp );
printf("%.2f%%", percentage);
return 0;
}
Sample input :
150 85
Sample output :
76.47%
but my output is :
-100.00%
Can someone help me fix this?
First of all, you should really read the values before using them:
scanf("%d %d", &sp, &bp);
percentage = ...
Secondly, this:
percentage = (sp-bp)/bp*100;
Does an integer division, so the rest of the division is totally discarded and you will end up with:
(150 - 85) / 85 == 0
0 * 100 == 0.
You should cast one end of the division to float before doing the operation:
percentage = (sp-bp)/(float)bp * 100;
Correct code:
#include <stdio.h>
int main() {
float percentage;
int sp;
int bp;
scanf("%d %d", &sp, &bp);
percentage = (sp-bp)/(float)bp * 100;
printf("%.2f%%", percentage);
return 0;
}
Explanation will add soon.
Code
#include <stdio.h>
int main()
{
float percentage=0;
int sp=0;
int bp=0;
scanf("%d%d", &sp, &bp );
//printf("\n%d%d", bp,sp);
percentage = (sp-bp)/(float)bp*100;
printf("%.2f%%", percentage);
}
Edit
Explain what i changed
As you scanning two integer with one scanf() function.
use scanf("%d%d", &sp, &bp );
Only remove space between two %d %d.
This white space is useless, because scanf() with %d format specifier consume any number of white space. but if you want to scan with %c use that.
scanf() consumes white space for all specifier except for %c, %n, %[…].
see more info on Jonathan Leffler answer
Initialize all variables like this (Declare and Define)
float percentage=0;
int sp=0;
int bp=0;
Because when we declare a variable inside function this is not initiate and have garbage value. This may cause undefined behavior.see C section
Suppose a situation that scanf() cant work properly and the integers sp and bp values remain with defaults. If we don't initiate them, garbage value remains.
Do declare and define at one time and initiate to zero. see
more on declare define
Add (float) explicit conversion to the division.
(sp-bp)/(float)bp*100;
In C result number of An integer that divides to another integer converted to integer and you lost decimal part because that is truncated.
see this
I want to get the command as input like
<char><number><number>
so is it possible to get it using one scanf() function in c ?
int i,j;
char c;
scanf("%c%d%d",&c, &i, &j);
I don't think that will work because the compiler will consider that as something together rather than one character, and two integers but to take the input in this format i.e. to make the compiler understand that you can take the entire line as a string and then parse the string to extract the character and the integers
char a[1000],c; int x,y;
scanf("%s",a);
c = a[0];
x = a[1] - '0';
y = a[2] - '0';
Also, #BLUEPIXY gave even better solution to your problem!
You can get more than 1 input in 1 scanf as the following:
char ch, string[100];
int number;
printf("--char--/--number--/--string--");
scanf(" %c%d %s",&ch,&number,&string[0]); // -> be careful about the whitespaces
printf("%c\n",ch);
printf("%d\n",number);
printf("%s\n",string);
You dont need to use spaces or \n to enter your input.
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();
}
Hi I am now learning the C language and I have a little problem with a exercise of the book I read. My code is this:
#include<stdio.h>
int main()
{
unsigned char one=0;
unsigned char two=0;
printf("Quantity 1 = ");
scanf("%d",&one);
printf("Quantity 2 = ");
scanf("%d",&two);
printf("The value is %d",one);
return 0;
}
Why when I am trying to see the value of one the initial value appears and not the value after the scanf?
You need to use int type in conjuction with %d specifier, and char with %c specifier. And %u with unsigned integers.
#include<stdio.h>
int main()
{
unsigned int one=0; unsigned int two=0;
printf("Quantity 1 = ");scanf("%u",&one);
printf("Quantity 2 = ");scanf("%u",&two);
printf("The value is %u",one);
return 0;
}
Basicaly, scanf will try to read integer from input and it will try to store it inside memory location that is not large enough, so you will have undefined behavior.
You can find good reference here.
However, if you try to use character for an input type, you may want ask yourself why you won't get a chance to enter a second Quantity (if you type 4 and press enter). This is because second scanf will read enter key as a character. Also, if you try to type 21 (for a twentyone), it will fill the first value with 2 and second with 1 (well, with their ASCII values).
So, be careful - be sure that you always choose the right type for your variables.
Never use scanf.
Never use scanf.
Seriously, never use scanf.
Use fgets (or getline, if you have it) to read an entire line of input from the user, then convert strings to numbers with strtol or its relatives strtod and strtoul. strsep may also be useful.
Check if scanf() is working properly by reading its return value. For quickstart, read the details about scanf() at this link.
What you are doing is inputting a integer using "%d" into an unsigned char variable, therefore scanf() may not be working as it should.
Change
unsigned char one=0; unsigned char two=0;
to
unsigned int one=0; unsigned int two=0;
and also use %u instead of %d then it will print the value after scanf().
You declared the variable one to be a char:
unsigned char one=0;
But then you told scanf to read an int:
scanf("%d",&one); /* %d means int */
Int is bigger than char (typically 4-bytes vs. 1-byte), causing the problem you describe.
Change your scanf to:
scanf("%c",&one); /* %c means char */
Then when you print out the value, also print a char:
printf("The value is %c",one); /* %c means char */