Reputation: 53
I need to get strings dynamically but as I need to get more than one string, I need to use functions. So far I wrote this (I put //**** at places i think might be wrong)
char* getstring(char *str);
int main() {
char *str;
strcpy(str,getstring(str));//*****
printf("\nString: %s", str);
return 0;
}
char* getstring(char str[]){//*****
//this part is copy paste from my teacher lol
char c;
int i = 0, j = 1;
str = (char*) malloc (sizeof(char));
printf("Input String:\n ");
while (c != '\n') {//as long as c is not "enter" copy to str
c = getc(stdin);
str = (char*)realloc(str, j * sizeof(char));
str[i] = c;
i++;
j++;
}
str[i] = '\0';//null at the end
printf("\nString: %s", str);
return str;//******
}
printf
in the function is working but not back in main
function.
I tried returning void
, getting rid of *s
or adding, making another str2
and tring to strcpy
there or not using strcpy
at all. Nothing seems to working. Am I misssing something? Or maybe this is not possible at all
//Thank you so much for your answers
Upvotes: 5
Views: 6402
Reputation: 11532
There are two things to take away from the lesson as it stands now:
(1) You should have one way of returning the reference to the new string, either as an argument passed by reference to the function OR as a return value; you should not be implementing both.
(2) Because the subroutine your teacher gave you allocates memory on the heap, it will be available to any part of your program and you do not have to allocate any memory yourself. You should study the difference between heap memory, global memory, and automatic (stack) memory so you understand the differences between them and know how to work with each type.
(3) Because the memory is already allocated on the heap there is no need to copy the string.
Given these facts, your code can be simplified to something like the following:
int main() {
char *str = getstring();
printf( "\nString: %s", str );
return 0;
}
char* getstring(){
.... etc
Going forward, you want to think about how you de-allocate memory in your programs. For example, in this code the string is never de-allocated. It is a good habit to think about your strategy for de-allocating any memory that you allocate.
Upvotes: 3
Reputation: 112
Let's simplify the code a bit:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* getstring()
{
char c = 0;
int i = 0, j = 2;
char *str = NULL;
if ((str = (char*) malloc(sizeof(char))) == NULL)
return NULL;
printf("Input String: ");
while (c = getc(stdin)) {
if (c == '\n') break;
str = (char*) realloc(str, j * sizeof(char));
str[i++] = c;
j++;
}
str[i] = '\0';
printf("getstring() String: %s\n", str);
return str;
}
int main()
{
char *str = getstring();
printf("main() String: %s\n", str);
free(str);
return 0;
}
Then execute:
$ make teststring && ./teststring
cc teststring.c -o teststring
Input String: asdfasfasdf
getstring() String: asdfasfasdf
main() String: asdfasfasdf
Upvotes: 1
Reputation: 726489
The problem is not with the function that you are using, but with the way you try copying its result into an uninitialized pointer.
Good news is that you don't have to copy - your function already allocates a string in dynamic memory, so you can copy the pointer directly:
char *str = getstring(str);
This should fix the crash. A few points to consider to make your function better:
main
needs to free(str)
when it is done in order to avoid memory leakrealloc
result in a temporary pointer, and do a NULL
check to handle out-of-memory situations properlyUpvotes: 3
Reputation: 30916
Getting the string part can be taken from this answer. Only put a \n
as input to the getline funtion.
char * p = getline('\n');
Three things :-
don't cast malloc, check if malloc/realloc
is successful and sizeof
is not a function.
Upvotes: 3