I have this function which receives a pointer to char array and initializes it to be len
repetitions of "a":
void test(char ** s, int len) {
*s = (char *)malloc(sizeof(char) * len);
int i;
for(i = 0; i < len; i++) {
*(s[i]) = 'a';
}
printf("%s\n", *s);
}
in the main()
I have this code:
char * s;
test(&s, 3);
but I get EXC_BAD_ACCESS (code=1, address=0x0)
error when I run main()
. The error occurs on the second iteration of the for loop in this line: *(s[i]) = 'a';
As far as I understand I'm not accessing the elements correctly, what is the correct way?
s
is declared as a pointer to a pointer. In reality, it's a pointer to a pointer to the start of an array, but that cannot be inferred from the type system alone. It could just as well be a pointer to a start of an array of pointers, which is how s[i]
treats it. You need to first derefence s
(to get the pointer to the array's start), and then index on it:
(*s)[i] = 'a';
Also, as @MFisherKDX correctly pointed out in comments, if you're going to pass *s
to printf
or any other standard string-manipulation function, you have to turn into a proper C string by terminating it with a 0 character.
This more clearly shows what you should be doing, while staying close to your original code:
void test(char ** s, int len) {
char *p = malloc(len);
int i;
for(i = 0; i < len; i++) {
p[i] = 'a';
}
printf("%s\n", p);
*s = p;
}
Note that sizeof(char)
is always one by definition, and in C there's no need to cast the result of malloc()
.
That code also has all your original problems in that it doesn't actually create a string that you can send to printf( "%s... )
. This fixes that problem:
void test(char ** s, int len) {
char *p = malloc(len+1);
int i;
for(i = 0; i < len; i++) {
p[i] = 'a';
}
p[i]='\0';
printf("%s\n", p);
*s = p;
}
And this is even easier, with no need to use a double- *
pointer:
char *test(int len) {
char *p = malloc(len+1);
int i;
for(i = 0; i < len; i++) {
p[i] = 'a';
}
p[i]='\0';
printf("%s\n", p);
return(p);
}
Instead of assignment
*(s[i]) = 'a';
use this:
(*s)[i] = 'a';
But do not forget that C/C++ strings are null terminated
Or you can use this approach to get more readable code:
void test(char** s, int len) {
// 1 char extra for zero
char *str = (char*)malloc(sizeof(char) * (len+1));
int i;
for(i = 0; i < len; i++)
str[i] = 'a';
// zero terminated string
str[len] = 0;
printf("%s\n", str);
*s = str;
}
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