Проверить scanf на ошибку

asked Nov 1, 2010 at 19:04

num will always contain an integer because it’s an int. The real problem with your code is that you don’t check the scanf return value. scanf returns the number of successfully read items, so in this case it must return 1 for valid values. If not, an invalid integer value was entered and the num variable did probably not get changed (i.e. still has an arbitrary value because you didn’t initialize it).

As of your comment, you only want to allow the user to enter an integer followed by the enter key. Unfortunately, this can’t be simply achieved by scanf("%dn"), but here’s a trick to do it:

int num;
char term;
if(scanf("%d%c", &num, &term) != 2 || term != 'n')
    printf("failuren");
else
    printf("valid integer followed by enter keyn");

answered Nov 1, 2010 at 19:42

You need to read your input as a string first, then parse the string to see if it contains valid numeric characters. If it does then you can convert it to an integer.

char s[MAX_LINE];

valid = FALSE;
fgets(s, sizeof(s), stdin);
len = strlen(s);
while (len > 0 && isspace(s[len - 1]))
    len--;     // strip trailing newline or other white space
if (len > 0)
{
    valid = TRUE;
    for (i = 0; i < len; ++i)
    {
        if (!isdigit(s[i]))
        {
            valid = FALSE;
            break;
        }
    }
}

answered Nov 1, 2010 at 19:08

John BodeJohn Bode

117k18 gold badges116 silver badges194 bronze badges

First ask yourself how you would ever expect this code to NOT return an integer:

int num; 
scanf("%d",&num);

You specified the variable as type integer, then you scanf, but only for an integer (%d).

What else could it possibly contain at this point?

answered Nov 1, 2010 at 19:38

I looked over everyone’s input above, which was very useful, and made a function which was appropriate for my own application. The function is really only evaluating that the user’s input is not a «0», but it was good enough for my purpose. Hope this helps!

#include<stdio.h>

int iFunctErrorCheck(int iLowerBound, int iUpperBound){

int iUserInput=0;
while (iUserInput==0){
    scanf("%i", &iUserInput);
    if (iUserInput==0){
        printf("Please enter an integer (%i-%i).n", iLowerBound, iUpperBound);
        getchar();
    }
    if ((iUserInput!=0) && (iUserInput<iLowerBound || iUserInput>iUpperBound)){
        printf("Please make a valid selection (%i-%i).n", iLowerBound, iUpperBound);
        iUserInput=0;
    }
}
return iUserInput;
}

answered Nov 2, 2014 at 17:53

Try this…

#include <stdio.h>

int main (void)
{
    float a;
    int q;

    printf("nInsert numbert");
    scanf("%f",&a);

    q=(int)a;
    ++q;

    if((q - a) != 1)
        printf("nThe number is not an integernn");
    else
        printf("nThe number is an integernn");

    return 0;
}

This is a more user-friendly one I guess :

#include<stdio.h>

/* This program checks if the entered input is an integer
 * or provides an option for the user to re-enter.
 */

int getint()
{
  int x;
  char c;
  printf("nEnter an integer (say -1 or 26 or so ): ");
  while( scanf("%d",&x) != 1 )
  {
    c=getchar();

    printf("You have entered ");
    putchar(c);
    printf(" in the input which is not an integer");

    while ( getchar() != 'n' )
     ; //wasting the buffer till the next new line

    printf("nEnter an integer (say -1 or 26 or so ): ");

  }

return x;
}


int main(void)
{
  int x;
  x=getint();

  printf("Main Function =>n");
  printf("Integer : %dn",x);

 return 0;
}

answered Mar 3, 2016 at 6:39

I developed this logic using gets and away from scanf hassle:

void readValidateInput() {

    char str[10] = { '' };

    readStdin: fgets(str, 10, stdin);
    //printf("fgets is returning %sn", str);

    int numerical = 1;
    int i = 0;

    for (i = 0; i < 10; i++) {
        //printf("Digit at str[%d] is %cn", i, str[i]);
        //printf("numerical = %dn", numerical);
        if (isdigit(str[i]) == 0) {
            if (str[i] == 'n')break;
            numerical = 0;
            //printf("numerical changed= %dn", numerical);
            break;
        }
    }
    if (!numerical) {
        printf("This is not a valid number of tasks, you need to enter at least 1 taskn");
        goto readStdin;
    }
    else if (str[i] == 'n') {
        str[i] = '';
        numOfTasks = atoi(str);
        //printf("Captured Number of tasks from stdin is %dn", numOfTasks);
    }
}

answered Jul 19, 2016 at 17:48

I’ve been searching for a simpler solution using only loops and if statements, and this is what I came up with. The program also works with negative integers and correctly rejects any mixed inputs that may contain both integers and other characters.

#include <stdio.h>
#include <stdlib.h> // Used for atoi() function
#include <string.h> // Used for strlen() function

#define TRUE 1
#define FALSE 0

int main(void)
{
    char n[10]; // Limits characters to the equivalent of the 32 bits integers limit (10 digits)
    int intTest;
    printf("Give me an int: ");

    do
    {        
        scanf(" %s", n);

        intTest = TRUE; // Sets the default for the integer test variable to TRUE

        int i = 0, l = strlen(n);
        if (n[0] == '-') // Tests for the negative sign to correctly handle negative integer values
            i++;
        while (i < l)
        {            
            if (n[i] < '0' || n[i] > '9') // Tests the string characters for non-integer values
            {              
                intTest = FALSE; // Changes intTest variable from TRUE to FALSE and breaks the loop early
                break;
            }
            i++;
        }
        if (intTest == TRUE)
            printf("%in", atoi(n)); // Converts the string to an integer and prints the integer value
        else
            printf("Retry: "); // Prints "Retry:" if tested FALSE
    }
    while (intTest == FALSE); // Continues to ask the user to input a valid integer value
    return 0;
}

answered Oct 4, 2014 at 1:34

This method works for everything (integers and even doubles) except zero (it calls it invalid):

The while loop is just for the repetitive user input. Basically it checks if the integer x/x = 1. If it does (as it would with a number), its an integer/double. If it doesn’t, it obviously it isn’t. Zero fails the test though.

#include <stdio.h> 
#include <math.h>

void main () {
    double x;
    int notDouble;
    int true = 1;
    while(true) {
        printf("Input an integer: n");
        scanf("%lf", &x);
        if (x/x != 1) {
            notDouble = 1;
            fflush(stdin);
        }
        if (notDouble != 1) {
            printf("Input is validn");
        }
        else {
            printf("Input is invalidn");
        }
        notDouble = 0;
    }
}

answered Jan 27, 2016 at 11:18

I was having the same problem, finally figured out what to do:

#include <stdio.h>
#include <conio.h>

int main ()
{
    int x;
    float check;
    reprocess:
    printf ("enter a integer number:");
    scanf ("%f", &check);
    x=check;
    if (x==check)
    printf("nYour number is %d", x);
    else 
    {
         printf("nThis is not an integer number, please insert an integer!nn");
         goto reprocess;
    }
    _getch();
    return 0;
}

answered Nov 9, 2013 at 11:26

I found a way to check whether the input given is an integer or not using atoi() function .

Read the input as a string, and use atoi() function to convert the string in to an integer.

atoi() function returns the integer number if the input string contains integer, else it will return 0. You can check the return value of the atoi() function to know whether the input given is an integer or not.

There are lot more functions to convert a string into long, double etc., Check the standard library «stdlib.h» for more.

Note : It works only for non-zero numbers.

#include<stdio.h>
#include<stdlib.h>

int main() {
    char *string;
    int number;

    printf("Enter a number :");
    string = scanf("%s", string);

    number = atoi(string);

    if(number != 0)
        printf("The number is %dn", number);
    else
        printf("Not a number !!!n");
    return 0;
}

answered Nov 8, 2016 at 5:29

asked Nov 1, 2010 at 19:04

num will always contain an integer because it’s an int. The real problem with your code is that you don’t check the scanf return value. scanf returns the number of successfully read items, so in this case it must return 1 for valid values. If not, an invalid integer value was entered and the num variable did probably not get changed (i.e. still has an arbitrary value because you didn’t initialize it).

As of your comment, you only want to allow the user to enter an integer followed by the enter key. Unfortunately, this can’t be simply achieved by scanf("%dn"), but here’s a trick to do it:

int num;
char term;
if(scanf("%d%c", &num, &term) != 2 || term != 'n')
    printf("failuren");
else
    printf("valid integer followed by enter keyn");

answered Nov 1, 2010 at 19:42

You need to read your input as a string first, then parse the string to see if it contains valid numeric characters. If it does then you can convert it to an integer.

char s[MAX_LINE];

valid = FALSE;
fgets(s, sizeof(s), stdin);
len = strlen(s);
while (len > 0 && isspace(s[len - 1]))
    len--;     // strip trailing newline or other white space
if (len > 0)
{
    valid = TRUE;
    for (i = 0; i < len; ++i)
    {
        if (!isdigit(s[i]))
        {
            valid = FALSE;
            break;
        }
    }
}

answered Nov 1, 2010 at 19:08

John BodeJohn Bode

117k18 gold badges116 silver badges194 bronze badges

First ask yourself how you would ever expect this code to NOT return an integer:

int num; 
scanf("%d",&num);

You specified the variable as type integer, then you scanf, but only for an integer (%d).

What else could it possibly contain at this point?

answered Nov 1, 2010 at 19:38

I looked over everyone’s input above, which was very useful, and made a function which was appropriate for my own application. The function is really only evaluating that the user’s input is not a «0», but it was good enough for my purpose. Hope this helps!

#include<stdio.h>

int iFunctErrorCheck(int iLowerBound, int iUpperBound){

int iUserInput=0;
while (iUserInput==0){
    scanf("%i", &iUserInput);
    if (iUserInput==0){
        printf("Please enter an integer (%i-%i).n", iLowerBound, iUpperBound);
        getchar();
    }
    if ((iUserInput!=0) && (iUserInput<iLowerBound || iUserInput>iUpperBound)){
        printf("Please make a valid selection (%i-%i).n", iLowerBound, iUpperBound);
        iUserInput=0;
    }
}
return iUserInput;
}

answered Nov 2, 2014 at 17:53

Try this…

#include <stdio.h>

int main (void)
{
    float a;
    int q;

    printf("nInsert numbert");
    scanf("%f",&a);

    q=(int)a;
    ++q;

    if((q - a) != 1)
        printf("nThe number is not an integernn");
    else
        printf("nThe number is an integernn");

    return 0;
}

This is a more user-friendly one I guess :

#include<stdio.h>

/* This program checks if the entered input is an integer
 * or provides an option for the user to re-enter.
 */

int getint()
{
  int x;
  char c;
  printf("nEnter an integer (say -1 or 26 or so ): ");
  while( scanf("%d",&x) != 1 )
  {
    c=getchar();

    printf("You have entered ");
    putchar(c);
    printf(" in the input which is not an integer");

    while ( getchar() != 'n' )
     ; //wasting the buffer till the next new line

    printf("nEnter an integer (say -1 or 26 or so ): ");

  }

return x;
}


int main(void)
{
  int x;
  x=getint();

  printf("Main Function =>n");
  printf("Integer : %dn",x);

 return 0;
}

answered Mar 3, 2016 at 6:39

I developed this logic using gets and away from scanf hassle:

void readValidateInput() {

    char str[10] = { '' };

    readStdin: fgets(str, 10, stdin);
    //printf("fgets is returning %sn", str);

    int numerical = 1;
    int i = 0;

    for (i = 0; i < 10; i++) {
        //printf("Digit at str[%d] is %cn", i, str[i]);
        //printf("numerical = %dn", numerical);
        if (isdigit(str[i]) == 0) {
            if (str[i] == 'n')break;
            numerical = 0;
            //printf("numerical changed= %dn", numerical);
            break;
        }
    }
    if (!numerical) {
        printf("This is not a valid number of tasks, you need to enter at least 1 taskn");
        goto readStdin;
    }
    else if (str[i] == 'n') {
        str[i] = '';
        numOfTasks = atoi(str);
        //printf("Captured Number of tasks from stdin is %dn", numOfTasks);
    }
}

answered Jul 19, 2016 at 17:48

I’ve been searching for a simpler solution using only loops and if statements, and this is what I came up with. The program also works with negative integers and correctly rejects any mixed inputs that may contain both integers and other characters.

#include <stdio.h>
#include <stdlib.h> // Used for atoi() function
#include <string.h> // Used for strlen() function

#define TRUE 1
#define FALSE 0

int main(void)
{
    char n[10]; // Limits characters to the equivalent of the 32 bits integers limit (10 digits)
    int intTest;
    printf("Give me an int: ");

    do
    {        
        scanf(" %s", n);

        intTest = TRUE; // Sets the default for the integer test variable to TRUE

        int i = 0, l = strlen(n);
        if (n[0] == '-') // Tests for the negative sign to correctly handle negative integer values
            i++;
        while (i < l)
        {            
            if (n[i] < '0' || n[i] > '9') // Tests the string characters for non-integer values
            {              
                intTest = FALSE; // Changes intTest variable from TRUE to FALSE and breaks the loop early
                break;
            }
            i++;
        }
        if (intTest == TRUE)
            printf("%in", atoi(n)); // Converts the string to an integer and prints the integer value
        else
            printf("Retry: "); // Prints "Retry:" if tested FALSE
    }
    while (intTest == FALSE); // Continues to ask the user to input a valid integer value
    return 0;
}

answered Oct 4, 2014 at 1:34

This method works for everything (integers and even doubles) except zero (it calls it invalid):

The while loop is just for the repetitive user input. Basically it checks if the integer x/x = 1. If it does (as it would with a number), its an integer/double. If it doesn’t, it obviously it isn’t. Zero fails the test though.

#include <stdio.h> 
#include <math.h>

void main () {
    double x;
    int notDouble;
    int true = 1;
    while(true) {
        printf("Input an integer: n");
        scanf("%lf", &x);
        if (x/x != 1) {
            notDouble = 1;
            fflush(stdin);
        }
        if (notDouble != 1) {
            printf("Input is validn");
        }
        else {
            printf("Input is invalidn");
        }
        notDouble = 0;
    }
}

answered Jan 27, 2016 at 11:18

I was having the same problem, finally figured out what to do:

#include <stdio.h>
#include <conio.h>

int main ()
{
    int x;
    float check;
    reprocess:
    printf ("enter a integer number:");
    scanf ("%f", &check);
    x=check;
    if (x==check)
    printf("nYour number is %d", x);
    else 
    {
         printf("nThis is not an integer number, please insert an integer!nn");
         goto reprocess;
    }
    _getch();
    return 0;
}

answered Nov 9, 2013 at 11:26

I found a way to check whether the input given is an integer or not using atoi() function .

Read the input as a string, and use atoi() function to convert the string in to an integer.

atoi() function returns the integer number if the input string contains integer, else it will return 0. You can check the return value of the atoi() function to know whether the input given is an integer or not.

There are lot more functions to convert a string into long, double etc., Check the standard library «stdlib.h» for more.

Note : It works only for non-zero numbers.

#include<stdio.h>
#include<stdlib.h>

int main() {
    char *string;
    int number;

    printf("Enter a number :");
    string = scanf("%s", string);

    number = atoi(string);

    if(number != 0)
        printf("The number is %dn", number);
    else
        printf("Not a number !!!n");
    return 0;
}

answered Nov 8, 2016 at 5:29