CODEWITHSUNDEEP
c
user2684719
user2684719

Reputation:

How to convert string to long long in c

How to convert string to long long 

int main() {
  char **ptr;
  long long val1 = strtoumax("1234567890123456",ptr, 10);
  printf("%ull\n",val1);
}

Its prints only 1015724736ll. How can I make it to print 1234567890123456

Upvotes: 1

Views: 2232

Answers (4)

nos
nos

Reputation: 229204

Use strtoll , strtoumax is for converting to a uintmax_t, which is usually not a long long.

Note that your printf format specifier is wrong too, there's no such thing as "%ull". "%ull" will get interpreted as the format specfier "%u" and an ordinary string of "ll". Passing an long long to be formatted as an unsigned int("%u") will likely give you unpredictable results.

You need to use 

 printf("%lld\n", val1);

If you had an unsigned long long, you'd use "%llu"

Upvotes: 4

chux
chux

Reputation: 154305

There are a number of issues.

Instead of printf("%ull\n",val1); use printf("%llu\n",val1);. Put "ll" first.

ptr should be char *ptr and then foo(..., &ptr, 10). (this prevents bad things as this parameters needs to point to memory you own.)

long long val1 should be unsigned long long val1 - you appear to want to use unsigned integers in the rest of the code

strtoumax should be strtoull - to match your destination type of unsigned long long.

Upvotes: 1

Yu Hao
Yu Hao

Reputation: 122443

You can use strtoll.

And for the format specifier in printf, use %ll, u stands for unsigned, which val1 isn't.

If unsigned long long is what you want, declare val1 as so, and use strtoull accordingly.

Upvotes: 0

jofel
jofel

Reputation: 3405

  1. use unsigned long long instead of long long

  2. use strtoull instead of strtoumax

  3. use printf("%llu\n",val1);

Upvotes: 2

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