#include <cstdio>
int main()
{
int i;
printf("%d", scanf("%d", &i));
}
Whatever number i input, i get the output:
1
Why is it so?
On success, the scanf function
returns the number of items successfully read.
This count can match the expected number of readings or fewer, even zero, if a matching failure happens.
In the case of an input failure before any data could be successfully read, EOF is returned.
Try this as well:
printf("%d",scanf("%d%d",&i,&i));
You output the result of scanf, which is not the number you enter, but the number of items that are successfully read. The number you enter is stored in i. To output it you would have to write an additional line:
#include <cstdio>
int main()
{
int i;
if (scanf("%d",&i) == 1)
printf("%d", i);
}
scanf() returns the number of items read when it succeeds. Here its reading only one number hence the output is 1 every time regardless of the number.
Related
why I get the result n as the length of first string
// Example program
#include <iostream>
#include <string>
int main()
{
int n = printf("jjj");
printf("%d",n); // jjj3
return 0;
}
thanks a lot
printf returns the number of characaters that have been written, as stated in its manual (printf(3))
Upon successful return, these functions return the number of characters printed (excluding the null byte used to end output to strings).
Hence the 3 in your output. The jjj printed string comes from the first printf call.
int n = printf("jjj"); // prints "jjj"
printf("%d", n); // prints "3" (assuming previous printf did not fail)
printf also return the number of char
#include<stdio.h>
#include<conio.h>
void main()
{
static int i=5;
clrscr();
if(--i)
{
printf("%d",i);
main();
}
getch();
}
When I am running this code, it is not giving any output and when I am removing getch() then after running when I am switching to output screen then it is showing ouput . Why?
I am using Turbo C++.
You would have seen output with debuger if you break on each iteration, but since you are obviously just executing your program what you see is the result of the last clrscr();. Since i is 0 you're not getting into the if, where the printing happens and you get right at the getch(). Hope that answers your question.
The printf function buffers output internally until either
the buffer is full
it's asked to print a newline (\n)
stdout is flushed with fflush()
You are not doing either of 2 or 3 and the buffer is surely bigger than 5 integers.
You probably want
printf("%d\n",i);
If you want the ints to be printed without a new line,
printf("%d ",i);
fflush(stdout);
I tried to solve this problem in UVa but I am getting a wrong answer and I cant seem to find the error
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2525
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
int t,j,k,i=1;
char a[1000];
while(scanf("%d",&t)!=EOF && t)
{
int sum=0;
getchar();
gets(a);
k=strlen(a);
for(j=0;j<k;j++)
{ if(a[j]=='a'||a[j]=='d'||a[j]=='g'||a[j]=='j'||a[j]=='m'||a[j]=='p'||a[j]=='t'||a[j]=='w'||a[j]==32)
sum=sum+1;
else if(a[j]=='b'||a[j]=='e'||a[j]=='h'||a[j]=='k'||a[j]=='n'||a[j]=='q'||a[j]=='u'||a[j]=='x')
sum=sum+2;
else if(a[j]=='c'||a[j]=='f'||a[j]=='i'||a[j]=='l'||a[j]=='o'||a[j]=='r'||a[j]=='v'||a[j]=='y')
sum=sum+3;
else if(a[j]=='s'||a[j]=='z')
sum=sum+4;
}
printf("Case #%d: %d\n",i,sum);
i++;
}
return 0;
}
In the problem description there is a single number that indicates the number of texts that will be in the input afterwards. Your original code was trying to read the number before every row of input.
The attempt to read the number in each one of the rows will fail since the input character set does not include any digits, so you could be inclined to think that there should be no difference. But there is, when you try to read a number it will start by consuming the leading whitespace. If the input is:
< space >< space >a
The output should be 3 (two '0' and one '2' keys), but the attempt to read the number out of the line will consume the two leading whitespace characters and the later gets will read the string "a", rather than " a". Your count will be off by the amount of leading whitespace.
separate your code into functions that do specific things: read the data from the file, calculate the number of key presses for each input, output the result
Benefit:
You can test each function independently. It is also easier to reason about the code.
The maximum size of an input is 100, this means you only need an array of 101 characters( including the final \0) for each input, not 1000.
Since this question is also tagged C++ try to use std::vector and std::string in your code.
The inner for seems right at a cursory glance. The befit of having a specialized function that computes the number of key presses is that you can easily verify it does the correct thing. Make sure you check it thoroughly.
I am confused about scanf's behaviour in the following program. scanf appears to input once, and then not input again, until a stream of characters is printed.
Below in a C program
#include<stdio.h>
int main()
{
int i, j=0;
do
{
++j;
scanf("%d", &i);
printf("\n\n%d %d\n\n", i, j);
}
while((i!=8) && (j<10));
printf("\nJ = %d\n", j);
return 0;
}
here, Till i am inputting any integer program works perfectly fine, but when a character is inputted it goes on printing the last inputed value of i and never stops(untill j is 10 when loop exits) for scanf to take next input.
output::
1 <-----1st input
1 1
2 <---- 2nd input
2 2
a <---- character input
2 3
2 4
2 5
2 6
2 7
2 8
2 9
2 10
J = 10
same thing is happening in c++ also.
#include<iostream>
using namespace std;
int main()
{
int i, j=0;
do
{
++j;
cin>>i;
cout<<i<<" "<<j<<"\n";
}
while((i!=8) && (j<10));
cout<<"\nj = "<<j<<"\n";
}
output of c++ program ::
1 <-----1st input
1 1
2 <-----2nd input
2 2
a <------ character input
0 3
0 4
0 5
0 6
0 7
0 8
0 9
0 10
j = 10
only change in c++ is that 0 is being printed instead of last value.
I know here integer values are expected by the program, but i want to know what happens when character is inputted in place of an integer?
what is the reason of all happening above?
When you enter a, then cin >> i fails to read it because the type of i is int to which a character cannot be read. That means, a remains in the stream forever.
Now why i prints 0 is a different story. Actually it can print anything. The content of i is not defined once the attempt to read fails. Similar thing happens with scanf as well.
The proper way to write it this:
do
{
++j;
if (!(cin>>i))
{
//handle error, maybe you want to break the loop here?
}
cout<<i<<" "<<j<<"\n";
}
while((i!=8) && (j<10));
Or simply this (if you want to exit loop if error occurs):
int i = 0, j = 0;
while((i!=8) && (j<10) && ( cin >> i) )
{
++j;
cout<<i<<" "<<j<<"\n";
}
If scanf sees a character in the input stream that doesn't match the conversion specifier, it stops the conversion and leaves the offending character in the input stream.
There are a couple of ways to deal with this. One is to read everything as text (using scanf with a %s or %[ conversion specifier or fgets) and then use atoi or strtol to do the conversion (my preferred method).
Alternately, you can check the return value of scanf; it will indicate the number of successful conversions. So, if scanf("%d", &i); equals 0, then you know there's a bad character in the input stream. You can consume it with getchar() and try again.
You can never expect your users to enter valid things. The best practice is to read the input into a string and try to convert it to integer. If the input is not an integer, you can give an error message to the user.
The problem is that when you enter an input that is not of the expected type (specified by %d for scanf, and the int type for cin>>i;, the inputstream is not advanced, which results in both operations trying to extract the same type of data from the exact same incorrect input (and failing just as well this time around too), thus you will never asked for another input.
To ensure this does not happen you will need to check the return value of both operations (read the manual for how each reports errors). If an error does happen (as when you enter a character), you will need to clear the error, consume the invalid input and try again. I find it better in C++ to read a whole line using std::gtline() instead of int or even std::string when geting input from ther user interactively, so you get into this "infinite" loop you experienced.
You are ignoring the return value. See what the manual says about scanf(3):
RETURN VALUE
These functions return the number of input items successfully matched and assigned, which can be fewer than provided for, or even zero in the event of an early matching failure.
It fails matching an integer.
You could check the return value of scanf to determine if an integer has been parsed correctly (return should =1). On failure, you have choices: either notify the user of the error and terminate, or recover by reading the next token with a scanf("%s" ...) perhaps with a warning.
For scanf, you need to check its return value to see if the conversion on the input worked. scanf will return the number of elements successfully scanned. If the conversion did not work, it will leave the input alone, and you can try to scan it differently, or just report an error. For example:
if (scanf("%d", &i) != 1) {
char buf[512];
fgets(buf, sizeof(buf), stdin);
printf("error in line %d: got %s", j, buf);
return 0;
}
In your program, since the input is left alone, your loop repeats trying to read the same input.
In C++, you check for failure using the fail method, but the input stream failure state is sticky. So it won't let you scan further without clearing the error state.
std::cin >> i;
if (std::cin.fail()) {
std::string buf;
std::cin.clear();
std::getline(cin, buf);
std::cout
<< "error in line " << j
<< ": got " << buf
<< std::endl;
return 0;
}
In your program, since you never clear the failure state, the loop repeats using cin in a failure state, so it just reports failure without doing anything.
In both cases, you might find it easier or more reliable to work with the input if you would read in the input line first, and then attempt to parse the input line. In pseudocode:
while read_a_line succeeds
parse_a_line
In C, the catch to reading a line is that if it is longer than your buffer, you will need to check for that and concatenate multiple fgets call results together to form the line. And, to parse a line, you can use sscanf, which is similar to scanf but works on strings.
if (sscanf(buf, "%d", &i) != 1) {
printf("error in line %d: got %s", j, buf);
return 0;
}
In C++, for quick low level parsing of formatted input, I also prefer sscanf. But, if you want to use the stream approach, you can convert the string buffer into a istringstream to scan the input.
std::getline(cin, buf);
if (std::cin.fail()) {
break;
}
std::istringstream buf_in(buf);
buf_in >> i;
if (buf_in.fail()) {
std::cout << "error in line " << j
<< ": got " << buf
<< std::endl;
return 0;
}
I'm an absolute beginner to programming and i'm just doing some exercises exercises for the beginning.
First of all, i'm using Visual C++ 2010 to compile C-Code. I just create a new project and choose an empty console application. After that, I create a ressource file named test.c and change in the file properties the elementype to C/C++ Compiler and compile as C++ Code, so that i can use #include <iostream> for the std::cin.get() command. Now the code:
#include <stdio.h>
#include <iostream>
int main()
{
int number1, number2;
int sum;
puts("Enter number 1 please:");
scanf_s("%d",&number1);
puts("Enter number 2 please:");
scanf_s("%d",&number2);
std::cin.get();
std::cin.get(); //(1)
sum = number1 + number2;
printf("The average is %f\n", sum/2);
return 0;
}
Now my problem ist that the "std::cin.get()" command is just ignored. Afer typing in the two numbers the program just stops and the console window closes.
Any idea where the problem is?
I have another question please.
Since my problem with holding the console open is solved (1), now my printf() gives me just zeros as output. I want to have a float number as output but no matter what i type in as number1 and number2 i always get "0.000000".
Since i'm still working on my little program to verify the input before it is accepted, i have another question please.
I want to use the following code just to check the input.
#include <stdio.h>
#include <iostream>
#include <ctype.h>
int main()
{
int number1, number2;
int sum;
puts("Enter number 1 please:");
scanf_s("%d",&number1);
if (isdigit(number1))
{
puts("Enter number 2 please:");
scanf_s("%d",&number2);
}
else
{
puts("Your input is not correct. Enter a number please.");
}
std::cin.get();
std::cin.get();
/*
sum = number1 + number2;
printf("The average is %f\n", sum/2); */
return 0;
}
Well it doensn't work. I type in a digit and my response is "Your input is not...". I have used the search and found the following: Check if User Inputs a Letter or Number in C. Unfortunately the suggestions doesn't help me.
It's not ignored. When you type your second number, then hit enter, it puts your number plus a newline character in the input stream. scanf removes the number, but leaves the newline character alone. When you call cin.get(), since there's a character in the stream, it doesn't wait for your input.
PigBen has already given you a good explanation of where you err. However, I have some additonla points to make about your program which won't fit into a comment:
You are mixing C and C++ input. Why? What's wrong with std::cin >> number1?
When you change number1 to double, you need to remember to change the formatting string in scanf(), too, while with IO streams the compiler will figure out everything for you. With streams and C++' strings, containers and other data structures, it's much harder to do something that compiles, but invokes the dreaded Undefined Behavior at run-time.
Also note that you do not check whether your inputting operations succeed. What happens if I invoke your program, and instead of passing it numbers, I enter non-digits? Never use input from users, files, or other externals sources unverified.
With IO streams, the input operator >> returns (a reference to) the stream, and you can use streams as if they were booleans, so you can do
if(std::cin >> number1)
// input succeeded
or
if( !(std::cin >> number2) ) // note the negation operator !
// input error
to check.
Streams enter a bad state internally after input/output errors. Any further IO operations will fail on a stream that had encountered an error. Therefore, if you want, you can delay input verification until all input operations are done:
std::cout << "Enter number 1 please:";
std::cin >> number1;
std::cout << "Enter number 2 please:";
std::cin >> number2;
if(!std::cin)
// input error
However, remember to always verify input before you first use it.
Note that I didn't check the output for errors. That's because it's hard to imagine something going wrong with output to the console. (And what would you do about it? Print an error message?) However, if you write into a file, remember to check output, too. It's easy for a file operation to go wrong.
In answer to your modified question it is because you are using ints for division. Change int sum to float sum and everything should be fine.
To answer your modified question: you're using the %f printf() format with an int, and that doesn't work. If you want to print out floating-point, you need to pass a double. You could print out (double)sum / 2 or even sum / 2.0, both of which yield doubles. (No, a float doesn't work the same for a variadic function like printf().) As it is, you're passing what is probably a four-byte type and telling printf() to treat it as an eight-byte type of different format, so it's no wonder you're not getting the expected results.
Alternately, you could switch to C++ iostreams, which save you the problem of matching types and knowing the default promotions. You'd still get an integer from sum/2, and that would drop any one-half, but it would be the right result.