Question

I’m trying to increase memory of a, but realloc doesnt seem to do anything. At 4th number program crashes. It seems also that numbers are put into a[0] even though counter is increased and should be a[counter]. I know I start at a[1], because I’m writing counter itself in a[0] when I’m done inputing.

Translation of printf: Input vector (you end input with any nonnumber character except dot).

#include <stdio.h>
#include <stdlib.h>

typedef float* vektor;

vektor beriVektor() {
    int counter = 0;
    float zacasna;
    float *a = (float *) malloc(sizeof(float));
    printf("Vpisi vektor (vpis zakljucis s katerim koli nestevilskim znakom razen pike):\n");
    while(scanf("%f", &zacasna)) {
        counter++;
        printf("%d\n", counter);
        printf("%d\n", sizeof(a));
        a = realloc(a, (sizeof(a) + sizeof(float)));
        a[counter] = zacasna;
    }
    if (sizeof(a) == sizeof(float)) {
        return NULL;
    }
    a[0] = counter;
    return a;
}

void izpisiVektor(vektor a) {
    if (a == NULL) {
        return;
    }
    else {
        int velikost = sizeof(a)/sizeof(float);
        for (int i = 0; i < velikost; i++) {
            printf("%f", *(a+i));
        }
    }

}

void main(){
    vektor a = beriVektor();
    izpisiVektor(a);

}

output:

ragezor@ragezor-VirtualBox:~$ ./dn09.o 
Vpisi vektor (vpis zakljucis s katerim koli nestevilskim znakom razen pike):
1 2 3 4
1
4
2
4
3
4
4
4
*** glibc detected *** ./dn09.o: realloc(): invalid next size: 0x09052008 ***
======= Backtrace: =========
/lib/libc.so.6(+0x6c501)[0x835501]
/lib/libc.so.6(+0x71c6d)[0x83ac6d]
/lib/libc.so.6(realloc+0xe3)[0x83af53]
./dn09.o[0x804850e]
./dn09.o[0x8048595]
/lib/libc.so.6(__libc_start_main+0xe7)[0x7dfce7]
./dn09.o[0x8048411]
======= Memory map: ========
001c9000-001e3000 r-xp 00000000 08:01 393295     /lib/libgcc_s.so.1
001e3000-001e4000 r--p 00019000 08:01 393295     /lib/libgcc_s.so.1
001e4000-001e5000 rw-p 0001a000 08:01 393295     /lib/libgcc_s.so.1
005d5000-005f1000 r-xp 00000000 08:01 393234     /lib/ld-2.12.1.so
005f1000-005f2000 r--p 0001b000 08:01 393234     /lib/ld-2.12.1.so
005f2000-005f3000 rw-p 0001c000 08:01 393234     /lib/ld-2.12.1.so
0069e000-0069f000 r-xp 00000000 00:00 0          [vdso]
007c9000-00920000 r-xp 00000000 08:01 393454     /lib/libc-2.12.1.so
00920000-00922000 r--p 00157000 08:01 393454     /lib/libc-2.12.1.so
00922000-00923000 rw-p 00159000 08:01 393454     /lib/libc-2.12.1.so
00923000-00926000 rw-p 00000000 00:00 0 
08048000-08049000 r-xp 00000000 08:01 140607     /home/ragezor/dn09.o
08049000-0804a000 r--p 00000000 08:01 140607     /home/ragezor/dn09.o
0804a000-0804b000 rw-p 00001000 08:01 140607     /home/ragezor/dn09.o
09052000-09073000 rw-p 00000000 00:00 0          [heap]
b7700000-b7721000 rw-p 00000000 00:00 0 
b7721000-b7800000 ---p 00000000 00:00 0 
b78d9000-b78da000 rw-p 00000000 00:00 0 
b78e7000-b78eb000 rw-p 00000000 00:00 0 
bfc2e000-bfc4f000 rw-p 00000000 00:00 0          [stack]
Aborted

edit:
Thank you. Very good answers from all of you.

Is there any way to find out how much memory space does have vektor a allocated?
Later in the code I check with sizeof(a)/sizeof(float) for number of elements in that array which now I understand is incorect. Luckily I have counter stored in a so i know how much elements i have, but if I wouldn’t have that information stored how would I know?

Answer 1

The realloc is incorrect. You are getting the size of a pointer on your machine, not the size of the allocated space so far.

float *a = (float *) malloc(sizeof(float));
/* .... */
a = realloc(a, (sizeof(a) + sizeof(float)));

So, suppose a float * occupies 4 bytes. You will always allocate 4 + sizeof(float) and eventually you’ll step outside. You need to keep track of the number of elements and then:

a = realloc(a, sizeof(float) * (el_no + 1));

Of course, a nicer form would be:

a = realloc(a, sizeof(*a) * (el_no + 1));

If you later decide to change the type of a you’ll be safe this way.

Edit

As a side note, calling realloc for each and every new element might seem like a good deal, but it’s slow. You should employ a strategy like “when I run out of space I will double the current used amount” or something in that line.