If optarg (the argument after the flag -s, from Getop library) isn't a digit, I want an error message to be printed and the program terminated, and if it is a digit, size needs to be set to optarg. The problem I have is while a command like the -s r will hit the error message, -s 2 will also, meaning it's interpreting the 2 as a string.
Debugging with for -s 2
#1 printf("%d",atoi(optarg));
#2 printf("%d",isdigit(atoi(optarg)));
I get the int value 2 for line #1,
and an int value of 0 for line #2.
So I was wondering why isdigit(atoi(optarg))) gives me a 0, when atoi(optarg) gives me an int. Is there a better way to check if optarg is an int?
int main (int argc, char *argv[]){
int size;
char option;
size = 0;
const char *optstring;
optstring = "rs:pih";
while ((option = getopt(argc, argv, optstring)) != EOF) {
switch (option) {
case 'r':
type_set = 1;
break;
**case 's':
capacity_set = 1;
if(isdigit(atoi(optarg))==0){
fprintf(stderr,"Argument after -s needs to be an int\n");
return 0;
}**
else{
size = atoi(optarg);
}
break;
default{
return 0;
}
isdigit takes a character and tells you if it is a digit. atoi takes a string (char *) and returns the number that the string represents. So when you call isdigit(atoi(... you're taking a number and treating it as a character. Since the charater codes for digits are 48..57, any number other than one of those will return false.
You probably want isdigit(*optarg) -- this will tell you if the first character of the argument (string) is a digit character. Of course, this only looks at the first character, so you might want isdigit(optarg[0]) && optarg[1] == 0 instead.
If you want to accept a number rather than a digit (and only a number), strtol works much better than atoi as it allows you to check for failures. Something like:
char *end;
errno = 0;
size = strtol(optarg, &end, 10);
while (isspace(*end)) ++end;
if (errno || *end) {
// an error occurred on conversion, or there is extra cruft
// after a number in the argument.
fprintf(stderr,"Argument after -s needs to be an int\n");
return 0; }
Chris Dodd has a great answer. Another, possibly simpler method is as follows ( assuming you just want to know if optarg is a SINGLE digit:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// Assuming ASCII. "string" must be one character long and it
// must be a digit
int isSingleDigit( char *str )
{
return( (strlen(str) == 1) && ( *str >= '0' && *str <= '9' ) );
}
int main (void)
{
char *str = "a";
char *str1 = "9";
char *str2 = "10";
char *str3 = "1a";
printf( "%s %d\n", str, isSingleDigit(str));
printf( "%s %d\n", str1, isSingleDigit(str1));
printf( "%s %d\n", str2, isSingleDigit(str2));
printf( "%s %d\n", str3, isSingleDigit(str3));
return 0;
}
Output:
a 0
9 1
10 0
1a 0
Related
I am trying to make put command line arguments by the user into an array but I am unsure how to approach it.
For example say I ran my program like this.
./program 1,2,3,4,5
How would I store 1 2 3 4 5 without the commas, and allow it to be passed to other functions to be used. I'm sure this has to do with using argv.
PS: NO space-separated, I want the numbers to parse into integers, I have an array of 200, and I want these numbers to be stored in the array as, arr[0] = 1, arr[1] = 2....
store 1 2 3 4 5 without the commas, and allow it to be passed to other functions to be used.
PS: NO space-separated, I want the numbers to parse into integers
Space or comma-separated doesn't matter. Arguments always come in as strings. You will have to do the work to turn them into integers using atoi (Ascii-TO-Integer).
Using spaces between arguments is the normal convention: ./program 1 2 3 4 5. They come in already separated in argv.
Loop through argv (skipping argv[0], the program name) and run them through atoi.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]) {
for(int i = 1; i < argc; i++) {
int num = atoi(argv[i]);
printf("%d: %d\n", i, num);
}
}
Using commas is going to make that harder. You first have to split the string using the kind of weird strtok (STRing TOKenizer). Then again call atoi on the resulting values.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[]) {
char *token = strtok(argv[1], ",");
while(token) {
int num = atoi(token);
printf("%d\n", num);
token = strtok(NULL, ",");
}
}
This approach is also more fragile than taking them as individual arguments. If the user types ./program 1, 2, 3, 4, 5 only 1 will be read.
One of the main disadvantages to using atoi() is it provides no check on the string it is processing and will happily accept atoi ("my-cow"); and silently fail returning 0 without any indication of a problem. While a bit more involved, using strtol() allows you to determine what failed, and then recover. This can be as simple or as in-depth a recovery as your design calls for.
As mentioned in the comment, strtol() was designed to work through a string, converting sets of digits found in the string to a numeric value. On each call it will update the endptr parameter to point to the next character in the string after the last digit converted (to each ',' in your case -- or the nul-terminating character at the end). man 3 strtol provides the details.
Since strtol() updates endptr to the character after the last digit converted, you check if nptr == endptr to catch the error when no digits were converted. You check errno for a numeric conversion error such as overflow. Lastly, since the return type is long you need to check if the value returned is within the range of an int before assigning to your int array.
Putting it altogether with a very minimal bit of error handling, you could do something like:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <errno.h>
#define NELEM 200 /* if you need a constant, #define one (or more) */
int main (int argc, char **argv) {
int arr[NELEM] = {0}, ndx = 0; /* array and index */
char *nptr = argv[1], *endptr = nptr; /* nptr and endptr */
if (argc < 2) { /* if no argument, handle error */
fputs ("error: no argument provided.\n", stderr);
return 1;
}
else if (argc > 2) { /* warn on more than 2 arguments */
fputs ("warning: more than one argument provided.\n", stdout);
}
while (ndx < NELEM) { /* loop until all ints processed or arr full */
int error = 0; /* flag indicating error occured */
long tmp = 0; /* temp var to hold strtol return */
char *onerr = NULL; /* pointer to next comma after error */
errno = 0; /* reset errno */
tmp = strtol (nptr, &endptr, 0); /* attempt conversion to long */
if (nptr == endptr) { /* no digits converted */
fputs ("error: no digits converted.\n", stderr);
error = 1;
onerr = strchr (endptr, ',');
}
else if (errno) { /* overflow in conversion */
perror ("strtol conversion error");
error = 1;
onerr = strchr (endptr, ',');
}
else if (tmp < INT_MIN || INT_MAX < tmp) { /* check in range of int */
fputs ("error: value outside range of int.\n", stderr);
error = 1;
onerr = strchr (endptr, ',');
}
if (!error) { /* error flag not set */
arr[ndx++] = tmp; /* assign integer to arr, advance index */
}
else if (onerr) { /* found next ',' update endptr to next ',' */
endptr = onerr;
}
else { /* no next ',' after error, break */
break;
}
/* if at end of string - done, break loop */
if (!*endptr) {
break;
}
nptr = endptr + 1; /* update nptr to 1-past ',' */
}
for (int i = 0; i < ndx; i++) { /* output array content */
printf (" %d", arr[i]);
}
putchar ('\n'); /* tidy up with newline */
}
Example Use/Output
This will handle your normal case, e.g.
$ ./bin/argv1csvints 1,2,3,4,5
1 2 3 4 5
It will warn on bad arguments in list while saving all good arguments in your array:
$ ./bin/argv1csvints 1,my-cow,3,my-cat,5
error: no digits converted.
error: no digits converted.
1 3 5
As well as handling completely bad input:
$ ./bin/argv1csvints my-cow
error: no digits converted.
Or no argument at all:
$ ./bin/argv1csvints
error: no argument provided.
Or more than the expected 1 argument:
$ ./bin/argv1csvints 1,2,3,4,5 6,7,8
warning: more than one argument provided.
1 2 3 4 5
The point to be made it that with a little extra code, you can make your argument parsing routine as robust as need be. While your use of a single argument with comma-separated values is unusual, it is doable. Either manually tokenizing (splitting) the number on the commas with strtok() (or strchr() or combination of strspn() and strcspn()), looping with sscanf() using something similar to the "%d%n" format string to get a minimal succeed / fail indication with the offset of the next number from the last, or using strtol() and taking advantage of its error reporting. It's up to you.
Look things over and let me know if you have questions.
This is how I'd deal with your requirement using strtol(). This does not damage the input string, unlike solutions using strtok(). It also handles overflows and underflows correctly, unlike solutions using atoi() or its relatives. The code assumes you want to store an array of type long; if you want to use int, you can add testing to see if the value converted is larger than INT_MAX or less than INT_MIN and report an appropriate error if it is not a valid int value.
Note that handling errors from strtol() is a tricky business, not least because every return value (from LONG_MIN up to LONG_MAX) is also a valid result. See also Correct usage of strtol(). This code requires no spaces before the comma; it permits them after the comma (so you could run ./csa43 '1, 2, -3, 4, 5' and it would work). It does not allow spaces before commas. It allows leading spaces, but not trailing spaces. These issues could be fixed with more work — probably mostly in the read_value() function. It may be that the validation work in the main loop should be delegated to the read_value() function — it would give a better separation of duty. OTOH, what's here works within limits. It would be feasible to allow trailing spaces, or spaces before commas, if that's what you choose. It would be equally feasible to prohibit leading spaces and spaces after commas, if that's what you choose.
#include <errno.h>
#include <limits.h>
#include <stdio.h>
#include <stdlib.h>
static int read_val(const char *str, char **eov, long *value)
{
errno = 0;
char *eon;
if (*str == '\0')
return -1;
long val = strtol(str, &eon, 0);
if (eon == str || (*eon != '\0' && *eon != ',') ||
((val == LONG_MIN || val == LONG_MAX) && errno == ERANGE))
{
fprintf(stderr, "Could not convert '%s' to an integer "
"(the leftover string is '%s')\n", str, eon);
return -1;
}
*value = val;
*eov = eon;
return 0;
}
int main(int argc, char **argv)
{
if (argc != 2)
{
fprintf(stderr, "Usage: %s n1,n2,n3,...\n", argv[0]);
exit(EXIT_FAILURE);
}
enum { NUM_ARRAY = 200 };
long array[NUM_ARRAY];
size_t nvals = 0;
char *str = argv[1];
char *eon;
long val;
while (read_val(str, &eon, &val) == 0 && nvals < NUM_ARRAY)
{
array[nvals++] = val;
str = eon;
if (str[0] == ',' && str[1] == '\0')
{
fprintf(stderr, "%s: trailing comma in number string\n", argv[1]);
exit(EXIT_FAILURE);
}
else if (str[0] == ',')
str++;
}
for (size_t i = 0; i < nvals; i++)
printf("[%zu] = %ld\n", i, array[i]);
return 0;
}
Output (program csa43 compiled from csa43.c):
$ csa43 1,2,3,4,5
[0] = 1
[1] = 2
[2] = 3
[3] = 4
[4] = 5
$
Hi folks thanks in advance for any help, I'm doing the CS50 course i'm at the very beginning of programming.
I'm trying to check if the string from the main function parameter string argv[] is indeed a number, I searched multiple ways.
I found in another topic How can I check if a string has special characters in C++ effectively?, on the solution posted by the user Jerry Coffin:
char junk;
if (sscanf(str, "%*[A-Za-z0-9_]%c", &junk))
/* it has at least one "special" character
else
/* no special characters */
if seems to me it may work for what I'm trying to do, I'm not familiar with the sscanf function, I'm having a hard time, to integrate and adapt to my code, I came this far I can't understand the logic of my mistake:
#include <cs50.h>
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
int numCheck(string[]);
int main(int argc, string argv[]) {
//Function to check for user "cooperation"
int key = numCheck(argv);
}
int numCheck(string input[]) {
int i = 0;
char junk;
bool usrCooperation = true;
//check for user "cooperation" check that key isn't a letter or special sign
while (input[i] != NULL) {
if (sscanf(*input, "%*[A-Za-z_]%c", &junk)) {
printf("test fail");
usrCooperation = false;
} else {
printf("test pass");
}
i++;
}
return 0;
}
check if the string from the main function parameter string argv[] is indeed a number
A direct way to test if the string converts to an int is to use strtol(). This nicely handles "123", "-123", "+123", "1234567890123", "x", "123x", "".
int numCheck(const char *s) {
char *endptr;
errno = 0; // Clear error indicator
long num = strtol(s, &endptr, 0);
if (s == endptr) return 0; // no conversion
if (*endptr) return 0; // Junk after the number
if (errno) return 0; // Overflow
if (num > INT_MAX || num < INT_MIN) return 0; // int Overflow
return 1; // Success
}
int main(int argc, string argv[]) {
// Call each arg[] starting with `argv[1]`
for (int a = 1; a < argc; a++) {
int success = numCheck(argv[a]);
printf("test %s\n", success ? "pass" : "fail");
}
}
sscanf(*input, "%*[A-Za-z_]%c", &junk) is the wrong approach for testing numerical conversion.
You pass argv to numcheck and test all strings in it: this is incorrect as argv[0] is the name of the running executable, so you should skip this argument. Note also that you should pass input[i] to sscanf(), not *input.
Furthermore, lets analyze the return value of sscanf(input[i], "%*[A-Za-z_]%c", &junk):
it returns EOF if the input string is empty,
it returns 0 if %*[A-Za-z_] fails,
it also returns 0 if the conversion %c fails after the %*[A-Za-z_] succeeds,
it returns 1 is both conversions succeed.
This test is insufficient to check for non digits in the string, it does not actually give useful information: the return value will be 0 for the string "1" and also for the string "a"...
sscanf() is very tricky, full of quirks and traps. Definitely not the right tool for pattern matching.
If the goal is to check that the strings contain only digits (at least one), use this instead, using the often overlooked standard function strspn():
#include <stdio.h>
#include <string.h>
int numCheck(char *input[]) {
int i;
int usrCooperation = 1;
//check for user "cooperation" check that key isn't a letter or special sign
for (i = 1; input[i] != NULL; i++) {
// count the number of matching character at the beginning of the string
int ndigits = strspn(input[i], "0123456789");
// check for at least 1 digit and no characters after the digits
if (ndigits > 0 && input[i][ndigits] == '\0') {
printf("test passes: %d digits\n", ndigits);
} else {
printf("test fails\n");
usrCooperation = 0;
}
}
return usrCooperation;
}
Let's try this again:
This is still your problem:
if (sscanf(*input, "%*[A-Za-z_]%c", &junk))
but not for the reason I originally said - *input is equal to input[0]. What you want to have there is
if ( sscanf( input[i], "%*[A-Za-z_]%c", &junk ) )
what you're doing is cycling through all your command line arguments in the while loop:
while( input[i] != NULL )
but you're only actually testing input[0].
So, quick primer on sscanf:
The first argument (input) is the string you're scanning. The type of this argument needs to be char * (pointer to char). The string typedef name is an alias for char *. CS50 tries to paper over the grosser parts of C string handling and I/O and the string typedef is part of that, but it's unique to the CS50 course and not a part of the language. Beware.
The second argument is the format string. %[ and %c are format specifiers and tell sscanf what you're looking for in the string. %[ specifies a set of characters called a scanset - %[A-Za-z_] means "match any sequence of upper- and lowercase letters and underscores". The * in %*[A-Za-z_] means don't assign the result of the scan to an argument. %c matches any character.
Remaining arguments are the input items you want to store, and their type must match up with the format specifier. %[ expects its corresponding argument to have type char * and be the address of an array into which the input will be stored. %c expects its corresponding argument (in this case junk) to also have type char *, but it's expecting the address of a single char object.
sscanf returns the number of items successfully read and assigned - in this case, you're expecting the return value to be either 0 or 1 (because only junk gets assigned to).
Putting it all together,
sscanf( input, "%*[A-Za-z_]%c", &junk )
will read and discard characters from input up until it either sees the string terminator or a character that is not part of the scanset. If it sees a character that is not part of the scanset (such as a digit), that character gets written to junk and sscanf returns 1, which in this context is treated as "true". If it doesn't see any characters outside of the scanset, then nothing gets written to junk and sscanf returns 0, which is treated as "false".
EDIT
So, chqrlie pointed out a big error of mine - this test won't work as intended.
If there are no non-letter and non-underscore characters in input[i], then nothing gets assigned to junk and sscanf returns 0 (nothing assigned). If input[i] starts with a letter or underscore but contains a non-letter or non-underscore character later on, that bad character will be converted and assigned to junk and sscanf will return 1.
So far so good, that's what you want to happen. But...
If input[i] starts with a non-letter or non-underscore character, then you have a matching failure and sscanf bails out, returning 0. So it will erroneously match a bad input.
Frankly, this is not a very good way to test for the presence of "bad" characters.
A potentially better way would be to use something like this:
while ( input[i] )
{
bool good = true;
/**
* Cycle through each character in input[i] and
* check to see if it's a letter or an underscore;
* if it isn't, we set good to false and break out of
* the loop.
*/
for ( char *c = input[i]; *c; c++ )
{
if ( !isalpha( *c ) && *c != '_' )
{
good = false;
break;
}
}
if ( !good )
{
puts( "test fails" );
usrCooperation = 0;
}
else
{
puts( "test passes" );
}
}
I followed the solution by the user "chux - Reinstate Monica". thaks everybody for helping me solve this problem. Here is my final program, maybe it can help another learner in the future. I decided to avoid using the non standard library "cs50.h".
//#include <cs50.h>
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <limits.h>
void keyCheck(int);
int numCheck(char*);
int main(int argc, char* argv[])
{
//Error code == 1;
int key = 0;
keyCheck(argc); //check that two parameters where sent to main.
key = numCheck(argv[1]); //Check for user "cooperation".
return 0;
}
//check for that main received two parameters.
void keyCheck(int key)
{
if (key != 2) //check that main argc only has two parameter. if not terminate program.
{
exit(1);
}
}
//check that the key (main parameter (argv [])) is a valid number.
int numCheck(char* input)
{
char* endptr;
errno = 0;
long num = strtol(input, &endptr, 0);
if (input == endptr) //no conversion is possible.
{
printf("Error: No conversion possible");
return 1;
}
else if (errno == ERANGE) //Input out of range
{
printf("Error: Input out of range");
return 1;
}
else if (*endptr) //Junk after numeric text
{
printf("Error: data after main parameter");
return 1;
}
else //conversion succesfull
{
//verify that the long int is in the integer limits.
if (num >= INT_MIN && num <= INT_MAX)
{
return num;
}
//if the main parameter is bigger than an int, terminate program
else
{
printf("Error key out of integer limits");
exit(1);
}
}
/* else
{
printf("Success: %ld", num);
return num;
} */
}
Im using sscanf to read/parse a 3 char string into an int as:
char strId[4] = "123";
int i;
int count = sscanf(strId, "%i" &i);
Im testing count==1 to check if the parse suceeds or fails.
"123" suceeds correctly - I want to consider this a number.
"aaa" fails correctly - I don't want to consider this a number.
but
"2aa" suceeds (count==1, i==2)- but I would like to identify this as a failure as I don't want to consider this a number.
How to simply parse strId to meet the above conditions?
Use strtol(3). It allows specification of an "end-of-parsing" pointer (endptr below). After the conversion has finished you can check wether or not it points at the end of string. If it doesn't, there were non-number characters left in the input.
long strtol(const char *restrict str, char **restrict endptr, int base);
From man strtol:
If endptr is not NULL, strtol() stores the address of the first
invalid character in *endptr. If there were no digits at all,
however, strtol() stores the original value of str in *endptr. (Thus,
if *str is not '\0' but **endptr is '\0' on return, the entire string
was valid.)
A sscanf() approach, use "%n".
"%n" saves the location where scanning stopped.
char strId[4] = "123";
int n = 0
sscanf(strId, "%i %n" &i, &n);
if (n == 0 || strId[n] != '\0') Handle_ProblemInput(strId);
This will pass: "123", " 123", "123 ".
And fail: "123a", "", " abc".
May/may not detect int overflow "12345678901234567890".
[Edit]
What is nice about the "%n" is that it can be used with complicated formats and with formats that otherwise end with fixed text. IOWs, simply detect if the scan made it all the way to '\0'?
int n = 0;
sscanf(buf, " %i %f %7s ,,, %c blah foo %n", ...);
if (n == 0 || buf[n] != '\0') Handle_ProblemInput(buf);
With strtol that's an easy task just do this
#include <stdio.h>
#include <stdlib.h>
char string[] = "123xyz";
char *endptr;
int value;
value = strtol(string, &endptr, 10);
if ((*string != '\0') && (*endptr == '\0'))
printf("%s is a number and has no unconvertible characters", string);
/* and now 'value' contains the integer value. */
If you want to use sscanf() then this should do it too
#include <stdio.h>
#include <string.h>
char string[] = "123";
int count;
int value;
int length;
length = strlen(string);
if ((sscanf(string, "%d%n", &value, &count) == 1) && (count == length))
printf("%s is a number and has no unconvertible characters", string);
/* and now 'value' contains the integer value. */
char str[255];
int count = sscanf((const char*)strId, "%i%254s", &i, str);
if it comes back with count == 1 it is okay?
I can have strings containing random 10 digit numbers e.g.
"abcgfg1234567890gfggf" or
"fgfghgh3215556890ddf" etc
basically any combination of 10 digits plus chars together in a string, so I need check the string to determine if a 10 digit number is present. I use strspn but it returns 0
char str_in[] = "abcgfg1234567890gfggf";
char cset[] = "1234567890";
int result;
result = strspn(str_in, cset); // returns 0 need it to return 10
The fact that the following code returns 0 instead of 10 highlights the problem. I asked this previously but most replies were for checking against a known 10 digit number. In my case the number will be random. Any better way than strspn?
It returns 0 because there are no digits at the start of the string.
The strspn() function calculates the length (in bytes) of the
initial segment of s which consists entirely of bytes in accept.
You need to skip non-digits - strcspn - and then call strspn on the string + that offset. You could try:
/* Count chars to skip. */
skip = strcspn(str_in, cset);
/* Measure all-digit portion. */
length = strspn(str_in + skip, cset)
EDIT
I should mention this must be done in a loop. For example if your string is "abcd123abcd1234567890" the first strspn will only match 3 characters and you need to look further.
Just use sscanf():
unsigned long long value;
const char *str_in = "abcgfg1234567890gfggf";
if(sscanf(str_in, "%*[^0-9]%uL", &value) == 1)
{
if(value >= 1000000000ull) /* Check that it's 10 digits. */
{
/* do magic here */
}
}
The above assumes that unsigned long long is large enough to hold a 10-digit decimal numbers, in practice this means it assumes that's a 64-bit type.
The %*[^0-9] conversion specifier tells sscanf() to ignore a bunch of initial characters that are not (decimal) digits, then convert an unsigned long long (%uL) directly after that. The trailing characters are ignored.
How about using a regex?
#include <stdio.h>
#include <stdlib.h>
#include <regex.h>
int
main(int argc, char **argv)
{
char str_in[] = "abcgfg1234567890gfggf";
int result = 0;
const char *pattern = "[0-9]{10}";
regex_t re;
char msg[256];
if (regcomp(&re, pattern, REG_EXTENDED|REG_NOSUB) != 0) {
perror("regcomp");
return(EXIT_FAILURE);
}
result = regexec(&re, str_in, (size_t)0, NULL, 0);
regfree(&re);
if (!result) {
printf("Regex got a match.\n");
} else if (result == REG_NOMATCH) {
printf("Regex got no match.\n");
} else {
regerror(result, &re, msg, sizeof(msg));
fprintf(stderr, "Regex match failed: %s\n", msg);
return(EXIT_FAILURE);
}
return(EXIT_SUCCESS);
}
strspn seems handy for this, but you would have to include it in a loop and search several times. Given the specific requirements, the easiest way is probably to make your own custom function.
int find_digits (const char* str, int n);
/* Searches [str] for a sequence of [n] adjacent digits.
Returns the index of the first valid substring containing such a sequence,
otherwise returns -1.
*/
#include <ctype.h>
int find_digits (const char* str, int n)
{
int result = -1;
int substr_len = 0;
int i = 0;
for(int i=0; str[i] != '\0'; i++)
{
if(isdigit(str[i]))
{
substr_len++;
}
else
{
substr_len=0;
}
if(substr_len == n)
{
result = i;
break;
}
}
return result;
}
(I just hacked this down here and now, not tested, but you get the idea. This is most likely the fastest algorithm for the task, that is, if performance matters at all)
Alternative use of sscanf()
(blatant variation of #unwind)
const char *str_in = "abcgfg0123456789gfggf";
int n1 = 0;
int n2 = 0;
// %*[^0-9] Scan any non-digits. Do not store result.
// %n Store number of characters read so far.
// %*[0-9] Scan digits. Do not store result.
sscanf(str_in, "%*[^0-9]%n%*[0-9]%n", &n1, &n2);
if (n2 == 0) return 0;
return n2 - n1;
Counts leading 0 characters as part of digit count.
Should one wish to avoid sscanf()
char str_in[] = "abcgfg1234567890gfggf";
const char *p1 = str_in;
while (*p1 && !isdigit(*p1)) p1++;
const char *p2 = p1;
while (isdigit(*p2)) p2++;
result = p2 - p1;
for testing a suit of "0123456789" inside a string you can do something like that:
int main()
{
char str_in[] = "abcgfg1234567890gfggf";
char cset[] = "1234567890";
int result;
int i;
int f;
i = 0;
f = 0;
while (str_in[i])
{
if (str_in[i] == cset[f])
{
f++;
if(f == strlen(cset))
return (f);
}
else
f = 0;
i++;
}
}
I am having trouble comparing two characters. I've written a very basic C problem to try out command line arguments.
Here is my code so far:
#include <stdio.h>
#include <unistd.h>
int main(int argc, char *argv[]) {
char ch;
char *type = "";
char *gender = "";
int baby = 0;
int count = 0;
/* Options:
* -t = type of pet
* -g = gender
* -b = baby or adult
*/
while ((ch = getopt(argc, argv, "t:g:b")) != EOF)
switch (ch) {
case 't':
type = optarg;
break;
case 'g':
gender = optarg;
break;
case 'b':
baby = 1;
break;
default:
fprintf(stderr, "Invalid option.\n");
return 1;
}
argc -= optind;
argv += optind;
printf("You have chosen a %s.\n", type);
if (gender == 'f')
puts("It's a girl");
if (gender == 'b')
puts("It's a boy.");
// The main command line arguments should be about the traits of the pet
printf("%s", "Traits: ");
for (count = 0; count < argc; count++)
printf("%s ", argv[count]);
return 0;
}
So if I type this into the terminal:
$ ./pet_shop -t dog -g f cute small
I get this as output:
You have chosen a dog:
Traits: cute small
The output it missing information about the gender, it should be a girl since I entered f. But I tried checking by printf("%i", gender) which gave the value 0. Is g == 'f' the incorrect way of comparing two characters?
gender is a char*, i.e. a pointer to a string's first charadcter. When you compare that to a single char, both the char and the pointer are converted to integers and an integer comparison is done.
To compare strings, use strcmp from <string.h>:
if (strcmp(gender, "f") == 0)
// it's a girl
Note the double quote (") which signifies a string, rather than a single character.
The problem is that you're comparing a string (or rather, a char*) to a char. This comparison (i.e. if(gender == 'f')) will compare the raw pointer value to the character instead of comparing the contents of the string to the character. Instead, you need to dereference the pointer and then compare that value, or index into the string, i.e. if(gender[0] == 'f').
Of course, it would also be a good idea to check that the string actually contains something before attempting that in order to avoid a segfault.
You have:
char *gender = "";
So gender is a string, not a character. To compare strings, use strcmp.
You first declared gender as a string:
char *gender = "";
Then you later treat is as a single character:
if(gender == 'f')
[...]
if(gender == 'b')
You need to clarify in your own mind what gender is, before you try and code it.
Pick one definition, and stick with it.