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

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);
}