How to convert string to int64_t?

A C99 conforming attempt.

[edit] employed @R. correction

// Note: Typical values of SCNd64 include "lld" and "ld".
#include <inttypes.h>
#include <stdio.h>

int64_t S64(const char *s) {
  int64_t i;
  char c ;
  int scanned = sscanf(s, "%" SCNd64 "%c", &i, &c);
  if (scanned == 1) return i;
  if (scanned > 1) {
    // TBD about extra data found
    return i;
    }
  // TBD failed to scan;  
  return 0;  
}

int main(int argc, char *argv[]) {
  if (argc > 1) {
    int64_t i = S64(argv[1]);
    printf("%" SCNd64 "\n", i);
  }
  return 0;
}

Users coming from a web search should also consider std::stoll.

It doesn't strictly answer this original question efficiently for a const char* but many users will have a std::string anyways. If you don't care about efficiency you should get an implicit conversion (based on the user-defined conversion using the single-argument std::string constructor) to std::string even if you have a const char*.

It's simpler than std::strtoll which will always require 3 arguments.

It should throw if the input is not a number, but see these comments.

There are a few ways to do it:

  strtoll(str, NULL, 10);

This is POSIX C99 compliant.

you can also use strtoimax; which has the following prototype:

 strtoimax(const char *str, char **endptr, int base);

This is nice because it will always work with the local intmax_t ... This is C99 and you need to include <inttypes.h>