'Fill a char ** in function argument lead to SIGSEVG [duplicate]
I want to fill a char ** variable from a function. When I used that filled value outside the filling function, I get an error.
code 139 (interrupted by signal 11: SIGSEGV)
Here is my filling function
int read(char ** tag_ids, int * nbtag)
{
......
// Loop through tags
*nbtag = tagCount;
tag_ids = (char**)malloc(tagCount * sizeof(char*));
for (idx = 0; idx < tagCount; idx++) {
error = readtag( idx, &tagData);
tag_ids[idx] = (char *)malloc(sizeof(char)* (tagData.epcLen+1));
if( tag_ids[idx] == NULL) {
printf("error \n");
}
strcpy(tag_ids[idx], epcStr);
printf("strcpy length %d data %s \n", tagData.epcLen, epcStr); // data OK
printf(" strcpy length %d data %s \n", tagData.epcLen, tag_ids[idx]); // data OK
}
return 0;
}
When I use that function:
char ** tagid = NULL;
int nbtag;
int error = read(tagid,&nbtag);
for (int i = 0 ; i<nbtag; i++){
printf("---> tag index : %d/%d \n", i, nbtag);
if( tagid == NULL) {
printf("NULL pointer \n"); // Detect pointer NULL why ???
}
if( tagid[i] == NULL) {
printf("NULL pointer \n");
}
printf("---> tag index : %d id %s \n", i, tagid[i]); // SIGSEGV
}
I think the char ** variable in argument of my function is a copy of my original variable but I don't really understand how to fill a char ** from function .
Solution 1:[1]
I think you can find a complete answer here, aniway if you want to modify the value of a char** variable and use it outside the functione, you need to use char*** passing argument by reference
Solution 2:[2]
Indeed, youre allocating a copy of the variable.
A good way of checking this, is printing &tag_ids outside and inside of the function. You'll see the values are different. If you want to allocate a double pointer inside a function, you need to pass a reference of the variable to be allocated. i.e.:
int func(char ***tag_ids, int *nbtag) {
*tag_ids = (char **)malloc(...);
...
}
and call the function like:
char **tagids = NULL;
int nbtag = 0;
int error = func(&tagids, &nbtag);
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Davide B |
| Solution 2 |
