Thread: Dynamic Memory Allocation

That whole thing is broke. Initially c is an unitialized pointer that you then try to store a value into; hence you are into undefined behavior. A malloc call only sizes an array once, using looping like this you would need an initial malloc call followed by realloc calls.

getch() is part of conio.h and is nonstandard. Just use getchar() here which is defined in the stdio header. If you are trying to make your own equivalent of getline here, you would want to look at a recursive solution; which for this case is much simpler to implement.

EDIT: Don't cast malloc. Casting Malloc-FAQ()

Also... you want to remember to null terminate the character array prior to printing it out.

Updated :

Code:

#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
int main()

{
    char *c, temp; int n = 0;
    c =  malloc(sizeof(char));

    printf("Enter a string:\n");
    while((temp=getche())!='\r')
    {
        *c[n]=temp;
        c[n+1]= malloc(sizeof(int));
        n++;
    }

    printf("%s",c);
    return 0;
}

getchar() wont work because the program is supposed to read and store as the user inputs the key.

Having an error on line 13: D:\Workspace\bitstr.c|13|error: invalid type argument of 'unary *' (have 'int')

Uncasting causes this to appear, as for line 14: D:\Workspace\bitstr.c|14|warning: assignment makes integer from pointer without a cast|. (I read the link, please clarify)

EDIT: hk_mp5kpdw, will fix that after I get the code to run at least.

Thank you all for helping.

Code:

c =  malloc(sizeof(char));

sizeof(char) is guaranteed to always be 1 (I believe)... so you could just say malloc(1) if you wanted. This allocates a single character from the heap and assigns the address to c. The only index that is valid from an array indexing scheme at this stage in the game is 0, for example c[0], anything else is invalid. Remember that for later.

Code:

char *c, temp; int n = 0;
...
while((temp=getche())!='\r')

getche returns an int not a char, this can be important in cases where you may expect to have EOF returned and need to process it.

Code:

*c[n]=temp;

The * part here makes no sense, c[n] by itself is perfectly fine. That's the cause of the error you mention in your post above.

Code:

c[n+1]= malloc(sizeof(int));

Remember what I said about c[0] being the only valid index (the first time through the loop)? Here you attempt to access c[1] which is an invalid access (trying to access memory you have not be given permission to access) and assign it an address off the heap. Also, c[1] a char and not a char* type so the assignment doesn't make sense from that perspective. Third, you are using malloc where I'm thinking you probably want to be using realloc within the loop. Fourth, sizeof(int)? The first time through the loop your allocated memory size is 1, you want to allocate 1 more byte than the previous iteration's size to increase the size.


How about a call to delete towards the end of your code just for grins?

Well, yes using malloc(1) would do the job, but still I used malloc(sizeof(char)) just to be sure.

About getche, it does return an int, but int can be used in char because of ASCII, but please clear out this, int takes up 4 bytes where as a character 1 byte, and as this is for string, would typecasting "(char)getche() " do the job?

c[n] is supposed to be pointing an address, therefore wouldn't "*c[n]" mean value at address in c[n]?
And about c[1] and others being invalid besides c[0], I tested it.
...And figured out that malloc wasn't necessary (verify please).

Code:

#include<stdio.h>
int main()
{
    char *c;
    c[0] = 'x';
    c[1] = 'y';
        printf("\n%c\n%c", c[0],c[1]);
}

I'm not much good at pointers and such stuff, so its kinda difficult for me to put my queries and thoughts here.
I'd really appreciate some more help because understanding this would mean a really great to me. Thank you.

Updated the code again, it ran after fixing some stuff up.

Code:

#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
char main()

{
    char *c, temp; int n = 0;
    c = malloc(1);
    printf("Enter a string:\n");
    while((temp=(char)getche())!='\r')
    {
     c[n] = temp;
     n++;
    }
    c[n]= '\0';
    printf("\n%s",c);
    return 0;
}

What I did come across while editing the code was that using "int *c, temp;" caused only the first character to show up while echoing the resulting string, obviously due to the difference in memory allocation that int and char took. Plus, I saw no difference in typecasting getche().


And, yeah, offtopic, manually adding [CODE] tags caused the code to appear only in plain text, why <-- I take that back

Originally Posted by slash.hack

About getche, it does return an int, but int can be used in char because of ASCII, but please clear out this, int takes up 4 bytes where as a character 1 byte, and as this is for string, would typecasting "(char)getche() " do the job?

You can't fit four bytes into one.

Code:

char c;

c = getchar();

However, that's not the correct way to do it, and there is a specific reason those functions return an int.

Code:

int c;
char key;

c = getchar();
if( c != EOF )
    key = c;

They return an int, because EOF is a negative integer that doesn't correspond to any char value. So if you squish EOF into a char, and you are trying to use that to read a file that has assorted bytes in it, you will get false positives.


Quzah.