How to check if particular characters appeared in same order? - c

Question

For example I have a string: PO-ELK=SAEER:SWE
- must always be before = and that should always be before :.
Is there a easy way I could check to see if those conditions are met in the string I input and if not an error is returned.

Answer 1

Just a little programming needed.
const char *symbol = "-=:";
const char *s = "PO-ELK=SAEER:SWE";
while (*s) {
if (*s == *symbol) symbol++;
s++;
}
if (*symbol) puts("Fail");
else puts("Success");

Answer 2

How about 3 calls to strchr()?
const char *s = "PO-ELK=SAEER:SWE";
const char *t;
if ((t = strchr(s, '-')) && (t = strchr(t, '=')) && (t = strchr(t, ':'))) puts("Success");
else puts("Failure");

Answer 3

Sort-of interesting problem, it might be good for code-golf
$ cat > yesno.c
#include <stdio.h>
#include <strings.h>
int main(int ac, char **av) {
char *s = "-=:",
*s0 = s,
*i = av[1];
while(*s && i && *i) {
if(index(s0, *i)) {
if(*i == *s) {
++i;
++s;
continue;
} else
break;
}
++i;
}
printf("%s\n", *s ? "no" : "yes");
}
^D
$ cc -Wall yesno.c
$ ./a.out PO-ELK=SAEER:SWE
There are some grey areas in the spec tho. Are the characters ever duplicated? And if so do we search for a subsequence that is in-order? Do we require that all three are found? Does the program need to be interactive or can it just use the shell args?

Answer 4

Use std::string::find. Make use of second argument.
First look for "-" and if it could be find, search for "=" passing position of "-" as start point. Then do the same for ":" with position of "=" as a second argument.

Answer 5

You can use the strchr function, which return a pointer to the character in the string. So you can call this function for each of the characters in question and check that the indexes are in the right order.
const char *str = "PO-ELK=SAEER:SWE";
const char *dashPtr = strchr(str,'-');
const char *equalPtr = strchr(str,'=');
const char *colonPtr = strchr(str,':');
if ((dashIdx == NULL) || (equalIdx == NULL) || (colonIdx == NULL)) {
printf("one of -, =, : not found\n");
} else {
if ((dashPtr < equalPtr) && (equalPtr < colonPtr)) {
printf("order correct\n");
} else {
printf("order incorrect\n");
}
}

Answer 6

Use state machine in this kind of problem :
Here is my solution, i didn't test it but it should give you some ideas
#include <string.h>
#include <stdio.h>
typedef enum {init, s1,s2,s3,end}eState;
int main()
{
char str[20] ="PO-ELK=SAEER:SWE";
int iExit = 0;
int i =0;
char c;
eState state = init;
while (!iExit)
{
c=str[i];
switch (state)
{
case init:
if (c =='-')
state = s1;
else if ((c =='=')||(c ==':'))
state = end;
break;
case s1:
if (c =='=')
state = s2;
else if(c ==':'||c=='-')
state = end;
break;
case s2:
if (c ==':')
state = s3;
else if(c =='='||c=='-')
state = end;
break;
case s3:
printf ("Succes \n");
iExit = 1;
break;
case end:
printf ("Error \n"),
iExit = 1;
break;
default :
break;
}
i++;
}
return 0;
}

Answer 7

You can use the idea shown in this demonstrative program. The advantage of this approach is that you can write just an if-statement to check a string.
#include <stdio.h>
#include <string.h>
int main( void )
{
const char *s = "PO-ELK=SAEER:SWE";
const char *t = "-=:";
size_t n = 0;
if ( s[n = strcspn( s + n, t )] == t[0] &&
s[n += 1 + strcspn( s + n + 1, t )] == t[1] &&
s[n += 1 + strcspn( s + n + 1, t )] == t[2] )
{
printf( "\"%s\" is a valid string\n", s );
}
else
{
printf( "\"%s\" is not a valid string\n", s );
}
s = "PO-ELK:SAEER=SWE";
n = 0;
if ( s[n = strcspn( s + n, t )] == t[0] &&
s[n += 1 + strcspn( s + n + 1, t )] == t[1] &&
s[n += 1 + strcspn( s + n + 1, t )] == t[2] )
{
printf( "\"%s\" is a valid string\n", s );
}
else
{
printf( "\"%s\" is not a valid string\n", s );
}
return 0;
}
The program output
"PO-ELK=SAEER:SWE" is a valid string
"PO-ELK:SAEER=SWE" is not a valid string
To guarantee that the string does not contain more than one of each target character you can write the condition the following way
if ( s[n = strcspn( s + n, t )] == t[0] &&
s[n += 1 + strcspn( s + n + 1, t )] == t[1] &&
s[n += 1 + strcspn( s + n + 1, t )] == t[2] &&
s[n += 1 + strcspn( s + n + 1, t )] == '\0' )