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);
}