Easiest way to convert string-of-integers to int

Question

I have a very long number with hundreds of digits which I have stored in a string. Now, I want to read it into an integer/long digit-by-digit.

long int strtol ( const char * str, char ** endptr, int base ); takes an end-pointer, but it doesn't seem like I can set it right, because it tries to read in all digits, and hence, returns LONG_MAX.

char str[]=
"70172427121883998797908792274921901699720888093776\
65727333001053367881220235421809751254540594752243\
52584907711670556013604839586446706324415722155397\
53697817977846174064955149290862569321978468622482\
83972241375657056057490261407972968652414535100474\
82166370484403199890008895243450658541227588666881\
16427171479924442928230863465674813919123162824586";

char *pEnd; //set pEnd to str[1]'s address how???
long num = strtol(str, &pEnd,  10);
cout << num << endl; //LONG_MAX

I thought of solutions like copying it to a new character array, and then applying strtol but that doesn't seem like the best way to do it.

So what would be the easiest/best way to do this?

Don't use the `endptr` in this way -- it's only for output. I suggest using something that's _not_ `strtol` if you actually want to extract individual digits. Like, um, casting to `char` and subtracting `'0'`! — Lightness Races in Orbit, Nov 28 '12 at 13:37
That's not what the endptr argument in strtol does. Read the manual page again. — Art, Nov 28 '12 at 13:37
@LightnessRacesinOrbit The solution shouldn't differ I think. — Anirudh Ramanathan, Nov 28 '12 at 13:37
@Cthulhu: Let the answerer decide that! These are two different languages and often have very different solutions. It's preferred to ask a question about one or the other, according to which you're _actually_ using. — Lightness Races in Orbit, Nov 28 '12 at 13:38
@LightnessRacesinOrbit Edited tags. C++. :) But with char arrays, I still think it should be relevant to both. — Anirudh Ramanathan, Nov 28 '12 at 13:39

score 6 · Accepted Answer · edited May 23 '17 at 12:04

6

If you want to read a single digit, you can simply use:

int digit = str[digitIndex] - '0';

This works, regardless of the character set used by the implementation, because the standard guarantees that the digits '0'-'9' will be represented by contiguous values.

The endptr parameter of strtol doesn't do what you think: It is used to return the position where the conversion stopped, not to tell the function where to stop.

If you want to convert a multi-digit substring of str, you can copy the required digits only to a new string, like this:

std::string partialStr(str+startIndex, numDigits);

and then convert it to an integer using one of the methods here.

edited May 23 '17 at 12:04

Community

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answered Nov 28 '12 at 13:36

interjay

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Oh damn. Forgot all about the ASCII values! Thanks. Any way to use strtol? – Anirudh Ramanathan Nov 28 '12 at 13:37
**If** I did want to read in more than one digit at a time, is there a better way? – Anirudh Ramanathan Nov 28 '12 at 13:40

score 4 · Answer 2 · answered Nov 28 '12 at 13:36

converting a single digit to its int representation is easy, you'd e.g. do

int i = str[0] - '0';

For the first digit. So you just need to create an array of integer, iterate over the string and convert each digit, storing the result in the array of ints.

score 1 · Answer 3 · answered Nov 28 '12 at 13:38

1

The end pointer is an output from that strtol. It tells you where it stopped reading, not the other way around.

If you want to read an individual digit then just do this:

digit = str[0] - '0';

answered Nov 28 '12 at 13:38

ams

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Alberto Bonsanto · Answer 4 · 2012-11-28T13:50:06.910

1

If the base is 10, and LENGTH is the string Length, do:

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

/*Prototype*/
long int intpow( int base, int exp);

int main(void)
{
    long int result = 0;
    int i, j = 0;
    char str[] = /* blah blah*/

    for( i = LENGHT - 1; i >= 0; i--){
        /* -48 because ASCII code. */
        result +=  ( str[ i ] - 48 ) * intpow( 10, j );
        j++;
    }
}

/* Recursive integer power function*/
long int intpow( int base, int exp)
{
    if( exp == 0)
      return 1;

    else
      return base * intpow( base, exp - 1);
}

edited Nov 28 '12 at 13:50

answered Nov 28 '12 at 13:44

Alberto Bonsanto

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Why didn't you use the `cmath` **pow** function? Also, aren't you missing the char to int conversion? – Anirudh Ramanathan Nov 28 '12 at 13:45
I know that i am editing, and pow function uses double parameters, he doesn't need that. – Alberto Bonsanto Nov 28 '12 at 13:48
I'll be using something like this. +1. Haven't tested the above though ;) – Anirudh Ramanathan Nov 28 '12 at 13:51

Arne Mertz · Answer 5 · 2012-11-29T08:42:46.657

1

use std::transform if you want to convert not only one of the digits but all of them:

size_t N = sizeof(str) -1 ; //or strlen(str) if str is a pointer, not an array
std::vector<int> theInts(N); 
std::transform(str, str+N, begin(theInts), [](char* pc){ return *pc -'0'; });

edited Nov 29 '12 at 08:42

answered Nov 28 '12 at 13:51

Arne Mertz

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The code isn't correct. **1.** it should be `str.size()`, not `sizeof(str)`, **2.** that `function` doesn't look right. But `std::transform` does have potential for converting multiple values, hence, +1. – Anirudh Ramanathan Nov 28 '12 at 14:29
**1.** In the original post it says `char str[] = "345335"` etc., so `str` is not a `std::string` but an array of chars that has no method `size()`. `sizeof(str)` will give the size of that array in bytes including the zero-delimiter, so I'll correct the code (`-1`). **2.** which `function` do you mean? If it's the fourth argument to `std::transform` that looks "not right" to you: it is a C++11 lambda, taking one parameter and returning a char. – Arne Mertz Nov 29 '12 at 08:42

Easiest way to convert string-of-integers to int

5 Answers5