# C++ Function 'stod': String-to-Double (Updated Version)

Hello, as promised I updated my stod-function from this thread ...

Enjoy !

``````double stod(const string &strInput)
{
/* STOD: String to double  */
/* Written by Mathias Van Malderen */
/* It's not allowed to sell this source code ! */

double dbl_one = 0;
double dbl_two = 0;
double dbl_final = 0;
int strlen;
bool dec_pt = false;
int nums_before_dec = 0;
int nums_after_dec = 0;

strlen = strInput.length();

/* Check whether the string can be transformed into a number */
if(strInput[0] == '0' && strInput[1] == '0')
{
// invalid number !
return 0;
}

for(int i = 0; i < strlen; i++)
{
if (strInput.c_str()[i] >= '0' && strInput.c_str()[i] <= '9')
{
// valid number !
} else if(strInput[i] == '.') {
if(dec_pt)
{
// there was already a decimal point counted
// invalid number !
return 0;
} else {
dec_pt = true; // increment by one
}
} else {
// invalid number !
return 0;
}
}

/* Convert the number */

// STEP 1: Calculate the amount of numbers before/after the decimal point (if there's one)
if(dec_pt) // if there's a decimal point in the number
{
for(int i = 0; i < strlen; i++)
{
if(strInput[i+1] != '.')
{
nums_before_dec++;
} else {
nums_before_dec++;
break;
}
}
nums_after_dec = strlen-nums_before_dec;
nums_after_dec -= 1;
} else {
nums_after_dec = 0;
nums_before_dec = strlen;
}

// STEP 2: Convert the string to a real number
for(int i = 0; i < nums_before_dec; i++)
{
dbl_one += (strInput[i] - '0') * apow(10, (nums_before_dec - i));
}

dbl_one = dbl_one / 10; // little fix

for(int i = 0; i < nums_after_dec; i++)
{
dbl_two += (strInput[i] - '0') / apow(10, i+1);
}

// STEP 3: Return the converted string as a double:
dbl_final = dbl_one + dbl_two;
return dbl_final;
}

/* This function 'apow' raises x to the power of y, it's a dependency of 'stod' */
double apow(const float &x, int y)
{
double result = 1;
if(y == 0)
return result;

if(y < 0)
{
y = -y;
for(int i = 0; i < y; i++)
result = result * x;
return 1/result;
}

for(int i = 0; i < y; i++)
result = result * x;

return result;
}``````
thoughtcoder 167

Here's a better implementation of apow, assuming I don't have bugs.

``````double apow(double x, int y)
{
if (y < 0)
return apow(1 / x, -y);
double m = x, p = 1.0;
while (y) {
if (y & 1)
p = p * m;
m = m * m;
y = y >> 1;
}
return p;
}``````
mvmalderen 2,072

I've rewritten my 'apow'-function, this is the final code:

``````double apow(const double &x, int y)
{
double result = 1;
if(y < 0)
return apow(1/x, -y);
for(int i = y; i > 0; i--)
result *= x;
return result;
}``````
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