CODEWITHSUNDEEP
ctypes
user2131316
user2131316

Reputation: 3277

how to check if the input is a number or not in C?

In the main function of C:

void main(int argc, char **argv)
{
   // do something here
}

In the command line, we will type any number for example 1 or 2 as input, but it will be treated as char array for the parameter of argv, but how to make sure the input is a number, in case people typed hello or c?

Upvotes: 22

Views: 260598

Answers (10)

shahab javanmardi
shahab javanmardi

Reputation: 43

The C library function int isdigit(int c) checks if the passed character is a decimal digit character.

#include <stdio.h>
#include <ctype.h>

int main () {
   int var1 = 'h';
   int var2 = '2';
    
   if( isdigit(var1) ) {
      printf("var1 = |%c| is a digit\n", var1 );
   } else {
      printf("var1 = |%c| is not a digit\n", var1 );
   }
   
   if( isdigit(var2) ) {
      printf("var2 = |%c| is a digit\n", var2 );
   } else {
      printf("var2 = |%c| is not a digit\n", var2 );
   }
   
   return(0);
}

the result is :

var1 = |h| is not a digit
var2 = |2| is a digit

Upvotes: 0

user9869932
user9869932

Reputation: 7337

This works for me

#include <string.h>

int isNumber(char *n) {

  int i = strlen(n);
  int isnum = (i>0);
  while (i-- && isnum) {
    if (!(n[i] >= '0' && n[i] <= '9')) {
      isnum = 0;
    }
  }
  return isnum;
}

e.g.:

printf("%i\n", isNumber("12"));   // 1
printf("%i\n", isNumber("033"));  // 1
printf("%i\n", isNumber("0"));    // 1
printf("%i\n", isNumber(""));     // 0
printf("%i\n", isNumber("aaa"));  // 0
printf("%i\n", isNumber("\n"));   // 0
printf("%i\n", isNumber("a0\n")); // 0

Upvotes: 0

Adrien.C
Adrien.C

Reputation: 209

Using scanf is very easy, this is an example :

if (scanf("%d", &val_a_tester) == 1) {
    ... // it's an integer
}

Upvotes: 7

mwjay
mwjay

Reputation: 1

if (sscanf(command_level[2], "%f%c", &check_f, &check_c)!=1)
{
        is_num=false;
}
else
{
        is_num=true;
}   

if(sscanf(command_level[2],"%f",&check_f) != 1) 
{
    is_num=false;
}

how about this?

Upvotes: 0

Jay
Jay

Reputation: 1309

I was struggling with this for awhile, so I thought I'd just add my two cents:

1) Create a separate function to check if an fgets input consists entirely of numbers: 

int integerCheck(){
char myInput[4];
fgets(myInput, sizeof(myInput), stdin);
    int counter = 0;
    int i;
    for (i=0; myInput[i]!= '\0'; i++){
        if (isalpha(myInput[i]) != 0){
            counter++;
            if(counter > 0){
                printf("Input error: Please try again. \n ");
                return main();
            }
        }

    }
    return atoi(myInput); 
}

The above starts a loop through every unit of an fgets input until the ending NULL value. If it comes across a letter or an operator, it adds "1" to the int "counter" which is initially set to 0. Once the counter becomes greater than 0, the nested if statement instructs the loop to print an error message & then restart the program. When the loops completes, if int 'counter' is still the value of 0, it returns the initially inputted integer to be used in the main function ...

2) the main function would be:

int main(void){
unsigned int numberOne;
unsigned int numberTwo;
numberOne = integerCheck();
numberTwo = integerCheck();
return numberOne*numberTwo;

}

Assuming both integers are inputted correctly, the example provided will yield the result of int "numberOne" multiplied by int "numberTwo". The program will repeat for however long it takes to get two properly inputted integers.

Upvotes: 0

Marius
Marius

Reputation: 2273

A self-made solution:

bool isNumeric(const char *str) 
{
    while(*str != '\0')
    {
        if(*str < '0' || *str > '9')
            return false;
        str++;
    }
    return true;
}

Note that this solution should not be used in production-code, because it has severe limitations. But I like it for understanding C-Strings and ASCII.

Upvotes: 5

ytoamn
ytoamn

Reputation: 377

The sscanf() solution is better in terms of code lines. My answer here is a user-build function that does almost the same as sscanf(). Stores the converted number in a pointer and returns a value called "val". If val comes out as zero, then the input is in unsupported format, hence conversion failed. Hence, use the pointer value only when val is non-zero.

It works only if the input is in base-10 form.

#include <stdio.h>
#include <string.h>
int CONVERT_3(double* Amt){

    char number[100];

    // Input the Data
    printf("\nPlease enter the amount (integer only)...");
    fgets(number,sizeof(number),stdin);

    // Detection-Conversion begins
    int iters = strlen(number)-2;
    int val = 1;
    int pos;
    double Amount = 0;
    *Amt = 0;
    for(int i = 0 ; i <= iters ; i++ ){
        switch(i){
            case 0:
                if(number[i]=='+'){break;}
                if(number[i]=='-'){val = 2; break;}
                if(number[i]=='.'){val = val + 10; pos = 0; break;}
                if(number[i]=='0'){Amount = 0; break;}
                if(number[i]=='1'){Amount = 1; break;}
                if(number[i]=='2'){Amount = 2; break;}
                if(number[i]=='3'){Amount = 3; break;}
                if(number[i]=='4'){Amount = 4; break;}
                if(number[i]=='5'){Amount = 5; break;}
                if(number[i]=='6'){Amount = 6; break;}
                if(number[i]=='7'){Amount = 7; break;}
                if(number[i]=='8'){Amount = 8; break;}
                if(number[i]=='9'){Amount = 9; break;}
            default:
                switch(number[i]){
                    case '.':
                        val = val + 10;
                        pos = i;
                        break;
                    case '0':
                        Amount = (Amount)*10;
                        break;
                    case '1':
                        Amount = (Amount)*10 + 1;
                        break;
                    case '2':
                        Amount = (Amount)*10 + 2;
                        break;
                    case '3':
                        Amount = (Amount)*10 + 3;
                        break;
                    case '4':
                        Amount = (Amount)*10 + 4;
                        break;
                    case '5':
                        Amount = (Amount)*10 + 5;
                        break;
                    case '6':
                        Amount = (Amount)*10 + 6;
                        break;
                    case '7':
                        Amount = (Amount)*10 + 7;
                        break;
                    case '8':
                        Amount = (Amount)*10 + 8;
                        break;
                    case '9':
                        Amount = (Amount)*10 + 9;
                        break;
                    default:
                        val = 0;
                }
        }
        if( (!val) | (val>20) ){val = 0; break;}// val == 0
    }

    if(val==1){*Amt = Amount;}
    if(val==2){*Amt = 0 - Amount;}
    if(val==11){
        int exp = iters - pos;
        long den = 1;
        for( ; exp-- ; ){
            den = den*10;
        }
        *Amt = Amount/den;
    }
    if(val==12){
        int exp = iters - pos;
        long den = 1;
        for( ; exp-- ; ){
            den = den*10;
        }
        *Amt = 0 - (Amount/den);
    }

    return val;
}


int main(void) {
    double AM = 0;
    int c = CONVERT_3(&AM);
    printf("\n\n%d    %lf\n",c,AM);

    return(0);
}

Upvotes: -1

Kranthi Kumar
Kranthi Kumar

Reputation: 1264

Another way of doing it is by using isdigit function. Below is the code for it: 

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
#define MAXINPUT 100
int main()
{
    char input[MAXINPUT] = "";
    int length,i; 

    scanf ("%s", input);
    length = strlen (input);
    for (i=0;i<length; i++)
        if (!isdigit(input[i]))
        {
            printf ("Entered input is not a number\n");
            exit(1);
        }
    printf ("Given input is a number\n");
}

Upvotes: 30

Neil Townsend
Neil Townsend

Reputation: 6084

Using fairly simple code:

int i;
int value;
int n;
char ch;

/* Skip i==0 because that will be the program name */
for (i=1; i<argc; i++) {
    n = sscanf(argv[i], "%d%c", &value, &ch);

    if (n != 1) {
        /* sscanf didn't find a number to convert, so it wasn't a number */
    }
    else {
        /* It was */
    }
}

Upvotes: 2

paxdiablo
paxdiablo

Reputation: 881463

You can use a function like strtol() which will convert a character array to a long.

It has a parameter which is a way to detect the first character that didn't convert properly. If this is anything other than the end of the string, then you have a problem.

See the following program for an example:

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

int main( int argc, char *argv[]) {
    int i;
    long val;
    char *next;

    // Process each argument given.

    for (i = 1; i < argc; i++) {
        // Get value with failure detection.

        val = strtol (argv[i], &next, 10);

        // Check for empty string and characters left after conversion.

        if ((next == argv[i]) || (*next != '\0')) {
            printf ("'%s' is not valid\n", argv[i]);
        } else {
            printf ("'%s' gives %ld\n", argv[i], val);
        }
    }

    return 0;
}

Running this, you can see it in operation:

pax> testprog hello "" 42 12.2 77x

'hello' is not valid
'' is not valid
'42' gives 42
'12.2' is not valid
'77x' is not valid

Upvotes: 17

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