segmentation fault on chars array [closed] - c

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I need to count how many strings there is in a "string list". Each string ends as usual with a NUL char ('\0'), and the list ends with two NUL chars in succession.
I wrote a function but I keep getting a segmentation fault:
int numStrsInList(const char* strList) {
int count = 0;
int flag = 0;
if(!(*strList))
return -1;
while (flag != 2) {
if (!(*strList)) {
count++;
flag++;
}
else
flag = 0;
strList++;
}
return count;
}
for example:
const char* empty = "\0";
const char* one = "Hell0 \t\n\v\f\rw0r1d\0";
const char* two = "Hello\0 \t\0";
const char* simple = "Hello\0world\0!\0";
the invocation for example:
numStrsInList(empty)
numStrsInList(one)
numStrsInList(two)
numStrsInList(simple)
for this strings the output should be:
0
1
2
3

Answer 1

There are several issues with your code....
int numStrsInList(const char* strList) {
int count = 0;
int flag = 0;
if(!(*strList)) // this is not right, numStrsInList("\0") returns -1 instead of 0
return -1; // did you mean if (!strlist) ??
while (flag != 2) {
if (!(*strList)) { // maybe using this notation if (!strlist[0])
count++; // would help in avoiding the error above
flag++; // c library has strlen() functions
} // that are much faster and will make your code more readable
else
flag = 0;
strList++;
}
return count;
}
}
Compare to, overall length added per request :)
int numStrsInList(const char* strList, int maxlen)
{
// returns the number of strings in a null terminated array of
// contiguous null-terminated strings.
// maxlen is the maximum overall length of the buffer,
// can be 0 to defeat length checking
const char* s;
int result = 0;
if (!strList) return -1;
for (s = strlist;
s > (char*)1 && s[0] != 0;
s = (maxlen) ? (memchr(s, 0, maxlen - (s - strlist)) + 1)
: (s + strlen(s) + 1) )
{
if ((s - strlist) > maxlen) return -1;
++result;
}
return result;
}

Answer 2

Just use standard C function strchr declared in the header <string.h>.
For example
#include <stdio.h>
#include <string.h>
size_t numStrsInList(const char *s)
{
size_t n = 0;
if (!(s[0] == '\0' && s[1] == '\0'))
{
do
{
s = strchr(s, '\0');
++n;
} while (*++s);
}
return n;
}
int main( void )
{
printf("The number of substrings is %zu\n", numStrsInList("\0"));
printf("The number of substrings is %zu\n", numStrsInList("Hell0 \t\n\v\f\rw0r1d\0")) ;
printf("The number of substrings is %zu\n", numStrsInList("Hello\0 \t\0"));
printf("The number of substrings is %zu\n", numStrsInList("Hello\0world\0!\0"));
}
The program output is
The number of substrings is 0
The number of substrings is 1
The number of substrings is 2
The number of substrings is 3
Without using the standard function strchr the function can be implemented the following way
size_t numStrsInList(const char *s)
{
size_t n = 0;
if (!(s[0] == '\0' && s[1] == '\0'))
{
do
{
while (*s) ++s;
++n;
} while (*++s);
}
return n;
}
Take into account that for example this string
"\0A\0"
contains two substrings: "" and "A". While this string "\0" contains neither substring.
As for your code then already this statement
if(!(*strList))
return -1;
does not make sense.
It seems you mean
if(!strList)
return -1;
that is that the pointer strList is not equal to NULL. However by analogy with standard string functions it is better when the caller checks whether the pointer is equal to NULL.