I'm making a calc function which is meant to check if the input is valid. So, I'll have 2 strings, one with what the user inputs (eg, 3+2-1 or maybe dog - which will be invalid), and one with the ALLOWED characters stored in a string, eg '123456789/*-+.^' .
I'm not sure how can I do this and have trouble getting it started. I know a few functions such as STRMCP, and the popular ones from the string.h file, but I have no idea how to use them to check every input.
What is the most simplest way to do this?
One way of proceeding is the following.
A string is an array of ascii codes. So if your string is
char formula[50];
then you have a loop
int n =0;
while (formula[n]!=0)
{
if ( (formula[n]<........<<your code here>> ))
{printf("invalid entry\n\n"); return -1; //-1 = error code
n++;
}
you need to put the logic into the loop, but you can test the ascii codes of each character with this loop.
There may be a more elegant way of solving this, but this will work if you put the correct conditional statement here to check the ascii code of each character.
The while statement checks to see ifyou got to the end of the string.
Here's a demonstration of how use strpbrk() to check all characters in a string are in your chosen set:
#include <string.h>
#include <stdio.h>
const char alphabet[] = "123456789/*+-=.^";
int main(void) {
const char a[] = "3+2-1";
const char b[] = "dog";
char *res = strpbrk(a, alphabet);
printf("%s %s\n", a, (res) ? "true" : "false");
res = strpbrk(b, alphabet);
printf("%s %s\n", b, (res) ? "true" : "false");
return 0;
}
That's not the fastest way to do this, but it's very easy to use.
However, if you are writing a calculator function, you really want to parse the string at the same time. A typical strategy would be to have two types of entity - operators (+-/*^) and operands (numbers, so -0.1, .0002, 42, etc). You would extract these from the string as you parse it, and just fail if you hit an invalid character. (If you need to handle parentheses, you'll need a stack for the parsing.... and you'll likely need to work with a stack anyway to process and evaluate the expression overall.)
Related
I have just started learning C after coding for some while in Java and Python.
I was wondering how I could "validate" a string input (if it stands in a certain criteria) and I stumbled upon the sscanf() function.
I had the impression that it acts kind of similarly to regular expressions, however I didn't quite manage to tell how I can create rather complex queries with it.
For example, lets say I have the following string:
char str[]={"Santa-monica 123"}
I want to use sscanf() to check if the string has only letters, numbers and dashes in it.
Could someone please elaborate?
The fact that sscanf allows something that looks a bit like a character class by no means implies that it is anything at all like a regular expression library. In fact, Posix doesn't even require the scanf functions to accept character ranges inside character classes, although I suspect that it will work fine on any implementation you will run into.
But the scanning problem you have does not require regular expressions, either. All you need is a repeated character class match, and sscanf can certainly do that:
#include <stdbool.h>
bool check_string(const char* s) {
int n = 0;
sscanf(s, "%*[-a-zA-Z0-9]%n", &n);
return s[n] == 0;
}
The idea behind that scanf format is that the first conversion will match and discard the longest initial sequence consisting of valid characters. (It might fail if the first character is invalid. Thanks to #chux for pointing that out.) If it succeeds, it will then set n to the current scan point, which is the offset of the next character. If the next character is a NUL, then all the characters were good. (This version returns OK for the empty string, since it contains no illegal characters. If you want the empty string to fail, change the return condition to return n && s[n] == 0;)
You could also do this with the standard regex library (or any more sophisticated library, if you prefer, but the Posix library is usually available without additional work). This requires a little bit more code in order to compile the regular expression. For efficiency, the following attempts to compile the regex only once, but for simplicity I left out the synchronization to avoid data races during initialization, so don't use this in a multithreaded application.
#include <regex.h>
#include <stdbool.h>
bool check_string(const char* s) {
static regex_t* re_ptr = NULL;
static regex_t re;
if (!re_ptr) regcomp((re_ptr = &re), "^[[:alnum:]-]*$", REG_EXTENDED);
return regexec(re_ptr, s, 0, NULL, 0) == 0;
}
I want to use sscanf() to check if the string has only letters, numbers and dashes in it.
Variation of #rici good answer.
Create a scanset for letters, numbers and dashes.
//v The * indicates to scan, but not save the result.
// v Dash (or minus sign), best to list first.
"%*[-0-9A-Za-z]"
// ^^^^^^ Letters a-z, both cases
// ^^^ Digits
Use "%n" to detect how far the scan went.
Now we can use determine if
Scanning stop due to a null character (the whole string is valid)
Scanning stop due to an invalid character
int n = 0;
sscanf(str, "%*[-0-9A-Za-z]%n", &n);
bool success = (str[n] == '\0');
sscanf does not have this functionality, the argument you are referring to is a format specifier and not used for validation. see here: https://www.tutorialspoint.com/c_standard_library/c_function_sscanf.htm
as also mentioned sscanf is for a different job. for more in formation see this link. You can loop over string using isalpha and isdigit to check if chars in string are digits and alphabetic characters or no.
char str[]={"Santa-monica 123"}
for (int i = 0; str[i] != '\0'; i++)
{
if ((!isalpha(str[i])) && (!isdigit(str[i])) && (str[i] != '-'))
printf("wrong character %c", str[i]);//this will be printed for spaces too
}
I want to ... check if the string has only letters, numbers and dashes in it.
In C that's traditionally done with isalnum(3) and friends.
bool valid( const char str[] ) {
for( const char *p = str; p < str + strlen(str); p++ ) {
if( ! (isalnum(*p) || *p == '-') )
return false;
}
return true;
}
You can also use your friendly neighborhood regex(3), but you'll find that requires a surprising amount of code for a simple scan.
After retrieving value on sscanf(), you may use regular expression to validate the value.
Please see Regular Expression ic C
I am trying to extract the number 4 and 3 from the string /ab/cd__my__sep__4__some__sep__3. I am trying with regex but not sure how would I do this. I wrote the following code, but it just prints out __my__sep__4__some__sep__3
#include <stdio.h>
#include <regex.h>
#include <string.h>
#include <stdlib.h>
int main() {
char* s = "/ab/cd__my__sep__4__some__sep__3";
regex_t regex;
int reti = regcomp(®ex,"__my__sep__([0-9]+)",REG_EXTENDED);
if(reti!=0) {
exit(-1);
}else {
regmatch_t match[2];
reti = regexec(®ex, s, 2, match, 0);
if(reti == 0) {
char *v = &s[match[1].rm_so];
ssize_t fl;
sscanf(v, "%zu", &fl);
printf("%s",v);
}else {
printf("else");
}
}
}
How could I extract the numbers 4 and 3 ?
match[0] refers to the part of the text matched by the entire pattern. match[1] is the match corresponding to the first capture (parenthesized subpattern).
Note that &s[match[1].rm_so] gives you a pointer to the start of the capture, but if you print the string at that point, you will get the part of the string starting at the beginning of the capture. In this case, that doesn't really matter. Since you're using sscanf to extract the integer value of the captured text, the fact that the substring isn't terminated immediately doesn't matter; it's not going to be followed by a digit, and sscanf will stop at the first non-digit.
But in the general case, it's possible that it will not be so easy to identify the end of the matched capture, and you can use one of these techniques:
If you want to print the capture, you can use a computed string width format: (See Note 1.)
printf("%.*s\n", match[1].rm_eo - match[1].rm_so, &s[match[1].rm_so]);
If you have strndup, you can easily create a dynamically-allocated copy of the capture: (See Note 2.)
char* capture = strndup(&s[match[1].rm_so], match[1].rm_eo - match[1].rm_so);
As a quick-and-dirty hack, it is also possible to just insert a NUL terminator (assuming that the searched string is not immutable, which means that it cannot be a string literal). You'll probably want to save the old value of the following character so that you can restore the string to it's original state:
char* capture = &s[match[1].rm_so];
char* rest = &s[match[1].rm_eo];
char saved_char = *rest;
*rest = 0;
/* capture now points to a NUL-terminated string. */
/* ... */
/* restore s */
*rest = saved_char;
None of the above is really necessary in the context of the original question, since the sscanf as written will work perfectly if you change the start of the string to scan from match[0] to match[1].
Notes:
In the general case, you should test to make sure that a capture was actually found before trying to use its offset. The rm_so member will be -1 if the capture was not found during the regex search That doesn't necessarily mean that the search failed, because the capture could be part of an alternative not used in the match.
Don't forget to free the copy when you no longer need it. If you don't have strndup, it's pretty easy to implement. But watch out for the corner cases.
Since you are using sscanf(), there is no need to use a regex. You can parse the two numbers from your string using sscanf() alone using the format string: "%*[^0-9]%d%*[^0-9]%d" where "%*[^0-9]" uses the assignment suppression '*' to read and discard all non-digit characters and then uses "%d" to extract the integer value. The full format-string just repeats those two patterns twice.
A short example using your input could be:
#include <stdio.h>
int main (void) {
char *s = "/ab/cd__my__sep__4__some__sep__3";
int a, b;
if (sscanf (s, "%*[^0-9]%d%*[^0-9]%d", &a, &b) == 2)
printf ("a: %d\nb: %d\n", a, b);
else {
fputs ("error: parse of integers failed.\n", stderr);
return 1;
}
}
Example Use/Output
$ ./bin/parse2ints
a: 4
b: 3
If you find yourself attempting to parse something that sscanf() cannot handle, then a regex is appropriate. Here, sscanf() is more than capable of handling your needs alone.
Create a regex format that only holds [0-9]. Then create a separate boolean function checking whether a character belongs or not to your regex. Then apply the function to your string. If true, add the character to the string you want to output
My problem is: Input the string then replace the word that we want to change
For example: input: i love coke
word: coke
replace: pepsi
result: i love pepsi
But when i run this code it crashed. Can you help show me the mistake?
#include <stdio.h>
#include<string.h>
char replace(char s1[100],char s2[100],char s3[100])
{
int k,i,j;
for(i=0;i<strlen(s1);i++)
for(j=0;j<strlen(s2);j++)
for(k=0;k<strlen(s3);k++)
{
if(s1[i]==s2[j])
{
s1[i]=s3[k];
}
}
return s3;
}
int main()
{
char s1[100],s2[100],s3[100];
printf("input string: ");gets(s1);
printf("Find string: ");gets(s2);
printf("Replace: ");gets(s3);
printf("Result: %s",replace(s1,s2,s3));
return 0;
}
I suggest you use a 4th buffer to store the generated result. You won't be able to replace locally if the word to be replaced and the new word aren't the same length.
Also, you are comparing characters individually. Just because you found a c doesn't automatically mean you found coke and that you should replace it. You must check the entire word is there before replacing anything. Use strstr() to locate substrings inside a string.
In addition, your function is returning a char, it should return a string (char *).
Furthermore, there are plenty of examples online on how to write a function to replace words on a string, so lets not be reduntant. Google it.
Strings in C are null terminated e.g. "i love coke\0". The string length does not include the null terminator. Because of this you are overwriting the null terminator after the 'e' with the 'i' in "pepsi".
A quick hack to check if null terminating the string would help, is to memset s1, s2, and s3 to 0.
Your approach doesn't quite work. What you need to do is search the input string for the word you wish to replace. So, before you even start switching things around, you need to search for the whole word you wish to replace.
Once you find that word, you need to then put in your new word in it's place, and then start searching for the word again untill you finish your input string.
So, for pseudo code:
for i in input //for every letter
if input[i] != lookfor[0]
results[i] put input[i] into new "results" array
else // We might have found it.
for j in lookfor // Go through coke, one at a time
if input[i+j] != lookfor[j] "c,o,k,e"
break; //You didn;t find coke, just "c" or "co" or "cok"
// If you got all the way through, you found coke.
//So now you have to switch that out for the new that in the result
results[i] = "pepsi" //Just be careful here, because this has a different index than i, because pespi's length != coke's length
Did that make sense?
First of all, your replace function is returning to char instead of char*. You can also define your function's return type as void and can make it return to char* buffer (in/out) parameter after in-function string operations. Moreover, you can use strtok(), strcmp() and strstr() predefined string.h functions to accomplish any kind of string operations.
Check this out to get information about standart string operation functions: String Operations
I know it's a little unorthodox and will probably cost me some downvotes, but since it's due in 1 hour and I have no idea where to begin I thought I'd ask you guys.
Basically I'm presented with a string that contains placeholders in + form, for example:
1+2+5
I have to create a function to print out all the possibilities of placing different combinations of any given series of digits. I.e. for the series:
[9,8,6] // string array
The output will be
16265
16285
16295
18265
18285
18295
19265
19285
19295
So for each input I get (number of digits)^(number of placeholders) lines of output.
Digits are 0-9 and the maximum form of the digits string is [0,1,2,3,4,5,6,7,8,9].
The original string can have many placeholders (as you'd expect the output can get VERY lengthly).
I have to do it in C, preferably with no recursion. Again I really appreciate any help, couldn't be more thankful right now.
If you can offer an idea, a simplified way to look at solving this, even in a different language or recursively, it'd still be ok, I could use a general concept and move on from there.
It prints them in different order, but it does not matter. and it's not recursive.
#include <stdlib.h>
#include <stdio.h>
int // 0 if no more.
get_string(char* s, const char* spare_chr, int spare_cnt, int comb_num){
for (; *s; s++){
if (*s != '+') continue;
*s = spare_chr[comb_num % spare_cnt];
comb_num /= spare_cnt;
};
return !comb_num;
};
int main(){
const char* spare_str = "986";
int num = 0;
while (1){
char str[] = "1+2+5";
if (!get_string(str, spare_str, strlen(spare_str), num++))
break; // done
printf("str num %2d: %s\n", num, str);
};
return 0;
};
In order to do the actual replacement, you can use strchr to find the first occurrence of a character and return a char * pointer to it. You can then simply change that pointer's value and bam, you've done a character replacement.
Because strchr searches for the first occurrence (before a null terminator), you can use it repeatedly for every value you want to replace.
The loop's a little trickier, but let's see what you make of this.
I've been trying to write a function in C that detects palindromes. The program currently looks like the following:
#include <stdio.h>
#include <string.h>
int main()
{
char palindrome[24]; int palength; int halflength;
gets(palindrome);
palength = strlen(palindrome);
halflength = palength / 2;
printf("That string is %u characters long.\r\n", palength);
printf("Half of that is %u.\r\n", halflength);
return 0;
}
Right now it detects the length of a string, and also shows what half of that is. This is just to make sure it is working how I think it should be. What the rest of the function should do (if possible) is take the integer from "halflength" and use that to take that amount of characters off of the beginning and end of the string and store those in separate strings. From there I'd be able to compare the characters in those, and be able return true or false if the string is indeed a palindrome.
TL;DR - Is it possible take a certain amount of characters (in this case the integer "halflength") off the front and end of a string and store them in separate variables. Read above for more information on what I'm trying to do.
P.S. - I know not to use gets(), but didn't feel like writing a function to truncate \n off of fgets().
int len = strlen(palindrome) - 1; // assuming no \n
int half = len << 1;
for (int i=0; i<=half; ++i)
if(palindrome[i] != palindrome[len-i])
return false;
return true;
What if you do something like this,
char *str1="lol",*str2;
str2=strrev(str1);
//if both are same then it actually is a palindrome ; )
Found out I was approaching the problem wrong. How I should be doing this is iterating over the characters using a pointer both backwards and forwards. Although if you'd still like to answer the original question it could still be useful at some point.