Traditional strtol() is usually used like this:
int main()
{
errno = 0;
char *s = "12345678912345678900";
char *endptr;
long i = strtol(s, &endptr, 10);
if(i == LONG_MAX && errno == ERANGE)
printf("overflow");
}
We need to access errno two times, and errno nowadays is usually a C macro finally expanded to a function. It seems a little expensive considering parsing string to integer isn't a heavy job.
So, is it better to implement strtol without errno but using some other ways to indicating overflow?
like:
long strtol(const char *nptr, char **endptr, int base, bool *is_overflow);
instead of
long strtol(const char *nptr, char **endptr, int base);
is it better to implement strtol without errno ...
No.
... but using some other ways to indicating overflow?
No.
long int strtol(const char * restrict nptr, char ** restrict endptr, int base);
strtol() is a standard C library function and any implementation must adhere to proper use of the 3 inputs and errno to be compliant.
Of course OP can implement some other my_strtol() as desired.
Any performance concerns around avoiding errno are a micro-optimization yet a reasonable design goal.
It really comes down to how to conveys problems of string to long
Overflow "12345678912345678901234567890"
No conversions "abc"
Excess junk "123 abc"
Leading space allowed, trailing space allowed?
Allow various bases?
Once functionality about all exceptional cases are defined, not just overflow, then coding concerns about errno is useful, even if unlikely to make any meaningful performance improvements.
IMO, coding to one base only is likely a more productive path to speed improvements than errno.
OP code is not a robust strtol() usage. Suggest:
char *s = "12345678912345678900";
char *endptr;
errno = 0;
long i = strtol(s, &endptr, 10);
if (errno == ERANGE) printf("Overflow %ld\n", i);
else if (s == endptr) printf("No conversion %ld\n", i);
else if (*endptr) printf("Extra Junk %ld\n", i);
else printf("Success %ld\n", i);
There is actually some overhead besides errno in strtol(), like skipping spaces, taking care of the base (10 or hexa), check characters ...
In a specific environment where speed is critical and you know the string provided is a number base 10 that fits in a long, you could make your own quick function, like
#include <ctype.h>
long mystrtol(char *s) {
long res = 0, minus = *s == '-';
if (minus || *s == '+') s++;
while (isdigit(*s)) {
res = res*10 + (*s++ - '0');
}
return minus ? -res : res;
}
and choose to inline it.
Related
So I have an assignment to figure out whether a number on the command line is either an integer or a double.
I have it mostly figured it out by doing:
sscanf(argv[x], "%lf", &d)
Where "d" is a double. I then cast it to an int and then subtract "d" with itself to check to see if it is 0.0 as such.
d - (int)d == 0.0
My problem is if the command line arguments contains doubles that can be technically classified as ints.
I need to classify 3.0 as a double whereas my solution considers it an int.
For example initializing the program.
a.out 3.0
I need it to print out
"3.0 is a double"
However right now it becomes
"3 is an int."
What would be a way to check for this? I did look around for similar problems which led me to the current solution but just this one edge case I do not know how to account for.
Thank you.
For example, a way like this:
#include <stdio.h>
int main(int argc, char *argv[]){
if(argc != 2){
puts("Need an argument!");
return -1;
}
int int_v, read_len = 0;
double double_v;
printf("'%s' is ", argv[1]);
//==1 : It was able to read normally.
//!argv[1][read_len] : It used all the argument strings.
if(sscanf(argv[1], "%d%n", &int_v, &read_len) == 1 && !argv[1][read_len])
puts("an int.");
else if(sscanf(argv[1], "%lf%n", &double_v, &read_len) == 1 && !argv[1][read_len])
puts("a double.");
else
puts("isn't the expected input.");
}
To test if a string will covert to a int and/or double (completely, without integer overflow, without undefined behavior), call strtol()/strtod(). #Tom Karzes
The trouble with a sscanf() approach is that the result is undefined behavior (UB) on overflow. To properly detect, use strtol()/strtod().
#include <errno.h>
#include <limits.h>
#include <stdbool.h>
#include <stdlib.h>
bool is_int(const char *src) {
char *endptr;
// Clear, so it may be tested after strtol().
errno = 0;
// Using 0 here allows 0x1234, octal 0123 and decimal 1234.
// or use 10 to allow only decimal text.
long num = strtol(src, &endptr, 0 /* or 10 */);
#if LONG_MIN < INT_MIN || LONG_MAX > INT_MAX
if (num < INT_MIN || num > INT_MAX) {
errno = ERANGE;
}
#endif
return !errno && endptr > src && *endptr == '\0';
}
bool is_double(const char *src) {
char *endptr;
// Clear, so it may be tested after strtod().
strtod(src, &endptr);
// In this case, detecting over/underflow IMO is not a concern.
return endptr > src && *endptr == '\0';
}
It is not entirely clear what the specific expectations are for your program, but it has at least something to do with the form of the input, since "3.0" must be classified as a double. If the form is all it should care about, then you should not try to convert the argument strings to numbers at all, for then you will run into trouble with unrepresentable values. In that case, you should analyze the character sequence of the argument to see whether it matches the pattern of an integer, and if not, whether it matches the pattern of a floating-point number.
For example:
int main(int argc, char *argv[]) {
for (int arg_num = 1; arg_num < argc; arg_num++) {
char *arg = argv[arg_num];
int i = (arg[0] == '-' || arg[0] == '+') ? 1 : 0; // skip any leading sign
// scan through all the decimal digits
while(isdigit(arg[i])) {
++i;
}
printf("Argument %d is %s.\n", arg_num, arg[i] ? "floating-point" : "integer");
}
}
That makes several assumptions, chief among them:
the question is strictly about form, so that the properties of your system's built-in data types (such as int and double) are not relevant.
each argument will have the form of either an integer or a floating-point number, so that eliminating "integer" as a possibility leaves "floating-point" as the only alternative. If "neither" is a possibility that must also be accommodated, then you'll also need to compare the inputs that do not have integer form to a pattern for floating-point numbers, too.
only decimal (or smaller radix) integers need be accommodated -- not, for example, hexadecimal inputs.
Under those assumptions, particularly the first, it is not just unnecessary but counterproductive to attempt to convert the arguments to one of the built-in numeric data types, because you would then come to the wrong conclusion about arguments that, say, are not within the bounds of representable values for those types.
For example, consider how the program should classify "9000000000". It has the form of an integer, but supposing that your system's int type has 31 value bits, that type cannot accommodate a value as large as the one the string represents.
int main (int argc,char *argv[])
{
if(argc==2)
{
int i;
double d;
d=atof(argv[1]);
i=atoi(argv[1]);
if(d!=i)
printf("%s is a double.",argv[1]);
else if(d==i)
printf("%s is an int.",argv[1]);
}
else
printf("Invalid input\n");
return 0;
}
You must add #include <stdlib.h>
Two part question;
I'm Coming from a high level Language, so this is a question about form not function;
I've written an isnumeric() function that takes a char[] and returns 1 if the string is a number taking advantage of the isdigit() function in ctype. Similar functions are builtin to other languages and I have always used something like that to integrity check the data before converting it to a numeric type. Mostly because some languages conversion functions fail badly if you try to convert a non-number string to an integer.
But it seems like a kludge having to do all that looping to compensate for the lack of strings in C, which poses the first part of the question;
Is it acceptable practice in C to trap for a 0 return from atoi() in lieu of doing an integrity check on the data before calling atoi()? The way atoi() (and other ascii to xx functions) works seems to lend itself well to eliminating the integrity check altogether. It would certainly seem more efficient to just skip the check.
The second part of the question is;
Is there a C function or common library function for a numeric integrity check
on a string? (by string, I of course mean char[])
Is it acceptable practice in C to trap for a 0 return from atoi() in lieu of doing an integrity check on the data before calling atoi()?
Never ever trap on error unless the error indicates a programming error that can't happen if there isn't a bug in the code. Always return some sort of error result in case of an error. Look at the OpenBSD strtonum function for how you could design such an interface.
The second part of the question is; Is there a C function or common library function for a numeric integrity check on a string? (by string, I of course mean char[])
Never use atoi unless you are writing a program without error checking as atoi doesn't do any error checking. The strtol family of functions allow you to check for errors. Here is a simply example of how you could use them:
int check_is_number(const char *buf)
{
const char *endptr;
int errsave = errno, errval;
long result;
errno = 0;
result = strtol(buf, &endptr, 0);
errval = errno;
errno = errsave;
if (errval != 0)
return 0; /* an error occured */
if (buf[0] == '\0' || *endptr != '\0')
return 0; /* not a number */
return 1;
}
See the manual page linked before for how the third argument to strtol (base) affects what it does.
errno is set to ERANGE if the value is out of range for the desired type (i.e. long). In this case, the return value is LONG_MAX or LONG_MIN.
If the conversion method returns an error indication (as distinct from going bananas if an error occurs, or not providing a definitive means to check if an error has occurred) then there is actually no need to check if a string is numeric before trying to convert it.
With that in mind, using atoi() is not a particularly good function to use if you need to check for errors on conversion. Zero will be returned for zero input, as well as an error, and there is no way to check on why. A better function to use is (assuming you want to read an integral value) is strtol(). Although strtol() returns zero on integer, it also returns information that can be used to check for failure. For example;
long x;
char *end;
x = strtol(your_string, &end, 10);
if (end == your_string)
{
/* nothing was read due to invalid character or the first
character marked the end of string */
}
else if (*end != '\0`)
{
/* an integral value was read, but there is following non-numeric data */
}
Second, there are alternatives to using strtol(), albeit involving more overhead. The return values from sscanf() (and, in fact, all functions in the scanf() family) can be checked for error conditions.
There is no standard function for checking if a string is numeric, but it can be easily rolled using the above.
int IsNumeric(char *your_string)
{
/* This has undefined behaviour if your_string is not a C-style string
It also deems that a string like "123AB" is non-numeric
*/
long x;
char *end;
x = strtol(your_string, &end, 10);
return !(end == your_string || *end != '\0`);
}
No (explicit) loops in any of the above options.
Is it acceptable practice in C to trap for a 0 return from atoi() in lieu of doing an integrity check on the data before calling atoi()?
No. #FUZxxl well answers that.
Is there a C function or common library function for a numeric integrity check on a string?
In C, the conversion of a string to a number and the check to see if the conversion is valid is usually done together. The function used depends on the type of number sought. "1.23" would make sense for a floating point type, but not an integer.
// No error handle functions
int atoi(const char *nptr);
long atol(const char *nptr);
long long atoll(const char *nptr);
double atof(const char *nptr);
// Some error detection functions
if (sscanf(buffer, "%d", &some_int) == 1) ...
if (sscanf(buffer, "%lf", &some_double) == 1) ...
// Robust methods use
long strtol( const char *nptr, char ** endptr, int base);
long long strtoll( const char *nptr, char ** endptr, int base);
unsigned long strtoul( const char *nptr, char ** endptr, int base);
unsigned long long strtoull( const char *nptr, char ** endptr, int base);
intmax_t strtoimax(const char *nptr, char ** endptr, int base);
uintmax_t strtoumax(const char *nptr, char ** endptr, int base);
float strtof( const char *nptr, char ** endptr);
double strtod( const char *nptr, char ** endptr);
long double strtold( const char *nptr, char ** endptr);
These robust methods use char ** endptr to store the string location where scanning stopped. If no numeric data was found, then *endptr == nptr. So a common test could is
char *endptr;
y = strto...(buffer, ..., &endptr);
if (buffer == endptr) puts("No conversion");
if (*endptr != '\0') puts("Extra text");
If the range was exceed these functions all set the global variable errno = ERANGE; and return a minimum or maximum value for the type.
errno = 0;
double y = strtod("1.23e10000000", &endptr);
if (errno == ERANGE) puts("Range exceeded");
The integer functions allow a radix selection from base 2 to 36. If 0 is used, the leading part of the string "0x", "0X", "0", other --> base 16, 16, 8, 10.
long y = strtol(buffer, &endptr, 10);
Read the specification or help page for more details.
You probably don't need a function to check whether a string is numeric. You will most likely need to convert the string to a number so just do that. Then check if the convertion is successful.
long number;
char *end;
number = strtol(string, &end, 10);
if ((*string == '\0') || (*end != '\0'))
{
// empty string or invalid number
}
the second argument of strtol is used to indicate where the parsing ended (the first non-numeric character). That character will be \0 if we've reached the end of the string. If you want to permit other characters after the number (like ), you can use switch to check for it.
strtol works with long integers. If you need some other type, you should consult the man page: man 3 strtol. For floating-point numbers you can use strtod.
Don't trap if the program logic permits that the string is not numeric (e.g. if it comes from the user or a file).
OP later commneted:
I'm looking for a way to determine if the string contains ONLY base 10 digits or a decimal or a comma. So if the string is 100,000.01 I want a positive return from func. Any other ascii characters anywhere in the string would result in a negative return value.
If is all your interest, use;
if (buffer[strspn(buffer, "0123456789.,")] == '\0') return 0; // Success
else return -1; // Failure
How can I convert a string to a long long in C?
I've got
char* example = "123";
I'd like to convert example to a long long so I'd want something like
long long n = example;
How can I do this?
Use the function strtoll:
#include <stdlib.h>
#include <errno.h>
char const * example = "123";
char * e;
errno = 0;
long long int n = strtoll(example, &e, 0);
if (*e != 0 || errno != 0) { /* error, don't use n! */ }
In fact, e will point to the next character after the converted sequence, so you can do even more sophisticated parsing with this. As it stands, we just check that the entire sequence has been converted. You can also inspect errno to see if an overflow occurred. See the manual for details.
(For historical interest: long long int and strtoll were introduced in C99. They're not available in C89/90. Equivalent functions strtol / strtoul / strtod exist, though.)
I tried
sscanf(str, "%016llX", &int64 );
but seems not safe. Is there a fast and safe way to do the type casting?
Thanks~
Don't bother with functions in the scanf family. They're nearly impossible to use robustly. Here's a general safe use of strtoull:
char *str, *end;
unsigned long long result;
errno = 0;
result = strtoull(str, &end, 16);
if (result == 0 && end == str) {
/* str was not a number */
} else if (result == ULLONG_MAX && errno) {
/* the value of str does not fit in unsigned long long */
} else if (*end) {
/* str began with a number but has junk left over at the end */
}
Note that strtoull accepts an optional 0x prefix on the string, as well as optional initial whitespace and a sign character (+ or -). If you want to reject these, you should perform a test before calling strtoull, for instance:
if (!isxdigit(str[0]) || (str[1] && !isxdigit(str[1])))
If you also wish to disallow overly long representations of numbers (leading zeros), you could check the following condition before calling strtoull:
if (str[0]=='0' && str[1])
One more thing to keep in mind is that "negative numbers" are not considered outside the range of conversion; instead, a prefix of - is treated the same as the unary negation operator in C applied to an unsigned value, so for example strtoull("-2", 0, 16) will return ULLONG_MAX-1 (without setting errno).
Your title (at present) contradicts the code you provided. If you want to do what your title was originally (convert a string to an integer), then you can use this answer.
You could use the strtoull function, which unlike sscanf is a function specifically geared towards reading textual representations of numbers.
const char *test = "123456789abcdef0";
errno = 0;
unsigned long long result = strtoull(test, NULL, 16);
if (errno == EINVAL)
{
// not a valid number
}
else if (errno == ERANGE)
{
// does not fit in an unsigned long long
}
At the time I wrote this answer, your title suggested you'd want to write an uint64_t into a string, while your code did the opposite (reading a hex string into an uint64_t). I answered "both ways":
The <inttypes.h> header has conversion macros to handle the ..._t types safely:
#include <stdio.h>
#include <inttypes.h>
sprintf( str, "%016" PRIx64, uint64 );
Or (if that is indeed what you're trying to do), the other way round:
#include <stdio.h>
#include <inttypes.h>
sscanf( str, "%" SCNx64, &uint64 );
Note that you cannot enforce widths etc. with the scanf() function family. It parses what it gets, which can yield undesired results when the input does not adhere to expected formatting. Oh, and the scanf() function family only knows (lowercase) "x", not (uppercase) "X".
gcc 4.4.4 c89
What is better to convert a string to an integer value.
I have tried 2 different methods atoi and sscanf. Both work as expected.
char digits[3] = "34";
int device_num = 0;
if(sscanf(digits, "%d", &device_num) == EOF) {
fprintf(stderr, "WARNING: Incorrect value for device\n");
return FALSE;
}
or using atoi
device_num = atoi(digits);
I was thinking that the sscanf would be better as you can check for errors. However, atoi doesn't doing any checking.
You have 3 choices:
atoi
This is probably the fastest if you're using it in performance-critical code, but it does no error reporting. If the string does not begin with an integer, it will return 0. If the string contains junk after the integer, it will convert the initial part and ignore the rest. If the number is too big to fit in int, the behaviour is unspecified.
sscanf
Some error reporting, and you have a lot of flexibility for what type to store (signed/unsigned versions of char/short/int/long/long long/size_t/ptrdiff_t/intmax_t).
The return value is the number of conversions that succeed, so scanning for "%d" will return 0 if the string does not begin with an integer. You can use "%d%n" to store the index of the first character after the integer that's read in another variable, and thereby check to see if the entire string was converted or if there's junk afterwards. However, like atoi, behaviour on integer overflow is unspecified.
strtol and family
Robust error reporting, provided you set errno to 0 before making the call. Return values are specified on overflow and errno will be set. You can choose any number base from 2 to 36, or specify 0 as the base to auto-interpret leading 0x and 0 as hex and octal, respectively. Choices of type to convert to are signed/unsigned versions of long/long long/intmax_t.
If you need a smaller type you can always store the result in a temporary long or unsigned long variable and check for overflow yourself.
Since these functions take a pointer to pointer argument, you also get a pointer to the first character following the converted integer, for free, so you can tell if the entire string was an integer or parse subsequent data in the string if needed.
Personally, I would recommend the strtol family for most purposes. If you're doing something quick-and-dirty, atoi might meet your needs.
As an aside, sometimes I find I need to parse numbers where leading whitespace, sign, etc. are not supposed to be accepted. In this case it's pretty damn easy to roll your own for loop, eg.,
for (x=0; (unsigned)*s-'0'<10; s++)
x=10*x+(*s-'0');
Or you can use (for robustness):
if (isdigit(*s))
x=strtol(s, &s, 10);
else /* error */
*scanf() family of functions return the number of values converted. So you should check to make sure sscanf() returns 1 in your case. EOF is returned for "input failure", which means that ssacnf() will never return EOF.
For sscanf(), the function has to parse the format string, and then decode an integer. atoi() doesn't have that overhead. Both suffer from the problem that out-of-range values result in undefined behavior.
You should use strtol() or strtoul() functions, which provide much better error-detection and checking. They also let you know if the whole string was consumed.
If you want an int, you can always use strtol(), and then check the returned value to see if it lies between INT_MIN and INT_MAX.
To #R.. I think it's not enough to check errno for error detection in strtol call.
long strtol (const char *String, char **EndPointer, int Base)
You'll also need to check EndPointer for errors.
Combining R.. and PickBoy answers for brevity
long strtol (const char *String, char **EndPointer, int Base)
// examples
strtol(s, NULL, 10);
strtol(s, &s, 10);
When there is no concern about invalid string input or range issues, use the simplest: atoi()
Otherwise, the method with best error/range detection is neither atoi(), nor sscanf().
This good answer all ready details the lack of error checking with atoi() and some error checking with sscanf().
strtol() is the most stringent function in converting a string to int. Yet it is only a start. Below are detailed examples to show proper usage and so the reason for this answer after the accepted one.
// Over-simplified use
int strtoi(const char *nptr) {
int i = (int) strtol(nptr, (char **)NULL, 10);
return i;
}
This is the like atoi() and neglects to use the error detection features of strtol().
To fully use strtol(), there are various features to consider:
Detection of no conversion: Examples: "xyz", or "" or "--0"? In these cases, endptr will match nptr.
char *endptr;
int i = (int)strtol(nptr, &endptr, 10);
if (nptr == endptr) return FAIL_NO_CONVERT;
Should the whole string convert or just the leading portion: Is "123xyz" OK?
char *endptr;
int i = (int)strtol(nptr, &endptr, 10);
if (*endptr != '\0') return FAIL_EXTRA_JUNK;
Detect if value was so big, the the result is not representable as a long like "999999999999999999999999999999".
errno = 0;
long L = strtol(nptr, &endptr, 10);
if (errno == ERANGE) return FAIL_OVERFLOW;
Detect if the value was outside the range of than int, but not long. If int and long have the same range, this test is not needed.
long L = strtol(nptr, &endptr, 10);
if (L < INT_MIN || L > INT_MAX) return FAIL_INT_OVERFLOW;
Some implementations go beyond the C standard and set errno for additional reasons such as errno to EINVAL in case no conversion was performed or EINVAL The value of the Base parameter is not valid.. The best time to test for these errno values is implementation dependent.
Putting this all together: (Adjust to your needs)
#include <errno.h>
#include <stdlib.h>
int strtoi(const char *nptr, int *error_code) {
char *endptr;
errno = 0;
long i = strtol(nptr, &endptr, 10);
#if LONG_MIN < INT_MIN || LONG_MAX > INT_MAX
if (errno == ERANGE || i > INT_MAX || i < INT_MIN) {
errno = ERANGE;
i = i > 0 : INT_MAX : INT_MIN;
*error_code = FAIL_INT_OVERFLOW;
}
#else
if (errno == ERANGE) {
*error_code = FAIL_OVERFLOW;
}
#endif
else if (endptr == nptr) {
*error_code = FAIL_NO_CONVERT;
} else if (*endptr != '\0') {
*error_code = FAIL_EXTRA_JUNK;
} else if (errno) {
*error_code = FAIL_IMPLEMENTATION_REASON;
}
return (int) i;
}
Note: All functions mentioned allow leading spaces, an optional leading sign character and are affected by locale change. Additional code is required for a more restrictive conversion.
Note: Non-OP title change skewed emphasis. This answer applies better to original title "convert string to integer sscanf or atoi"
If user enters 34abc and you pass them to atoi it will return 34.
If you want to validate the value entered then you have to use isdigit on the entered string iteratively