string to long, not giving me the correct answer - c

Question

I am trying to convert number stored in the c string to long int. But I am not getting the expected output:
char str[] = "987654321012345";
long int num ;
num = 0;
//num = atol(str);
num = strtol(str, (char **) NULL, 10);
printf("%ld", num);
Output : 821493369
gcc version 4.4.7 20120313 (Red Hat 4.4.7-16)
Can you please advise what am I doing wrong here? Thanks.

Answer 1

In addition to using long long, you can use exact width types from stdint.h. For instance, to guarantee and 64-bit signed number you can use the int64_t type. Regardless what you do, do not cast NULL to char ** and always validate your conversions. E.g.,
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
int main (void) {
char str[] = "987654321012345";
long num = 0;
errno = 0;
num = strtol (str, NULL, 10);
if (errno) { /* validate strtol conversion */
perror ("strtol conversion failed.");
return 1;
}
printf ("%ld\n", num);
return 0;
}
Example Use/Output
$ ./bin/strtoltst
987654321012345
There are additional error checks you can do on the conversion, but at minimum, make sure errno is not set following your call to strtol or strtoll.
If you would like to use the guaranteed width types, then you can make the following changes:
...
#include <stdint.h>
...
int64_t num = 0;
...
num = strtoll (str, NULL, 10);
The result is the same.

Answer 2

I could reproduce and fix.
Code reproducing the problem
#include <stdio.h>
//#include <stdlib.h>
int main()
{
char str[] = "987654321012345";
long int num ;
char *ix;
num = 0;
//num = atol(str);
num = strtol(str, &ix, 10);
printf("%ld\n", num);
printf("%lu\n", sizeof(long));
return 0;
}
Gives as expected:
821493369
8
After a compilation warning
warning: implicit declaration of strtoll
Reason: as strtoll is undeclared, it is assumed to return an int, so the long value is first truncate to an int and than promoted back to a long.
Fix: just uncomment the #include <stdlib.h> line...
Conclusion: warnings are not meant to be ignored!

Answer 3

You should use long long as datatype for your number.
char str[] = "987654321012345";
long long num = strtoll(str, (char **)NULL, 10);
printf("%lld", num);

Answer 4

According to the C data type:
Long signed integer type. Capable of containing at least the [−2,147,483,647, +2,147,483,647] range; thus, it is at least 32 bits in size.
Long unsigned integer type. Capable of containing at least the [0, 4,294,967,295] range;
Is not enough, so you need a long long or unsigned long long and use %lli or %llu

Answer 5

"987654321012345" is too large.
Strol is outputing long type variable.
Long value is –2,147,483,648 to 2,147,483,647.
try
char str[] = "987654321012345";
char *pEnd;
long long num;
num = 0;
num = strtoull(str,&pEnd, 10);
printf("%lld", num);
return 0;
long long instead of long
strtoull instead of strtol
%lld instead of %ld