I cannot read string after realloc - c

I have this function, I realocate memory, but when i want to read string it doesn't work=>error
Student is a Struct
void insertStudent(Student **myStudents, int *size)
{
int newSize = *size + 1;
*myStudents = (Student*)realloc(myStudents, newSize*sizeof(Student));
printf("Enter your grade: ");
scanf("%f", &(*myStudents)[*size - 1].nota);
printf("Enter your first name: ");
scanf("%s", &(*myStudents)[newSize-1].firstName);
printf("Enter your second name: ");
scanf("%s", &(*myStudents)[*size - 1].lastName);
//Generate a new code
/*int code = rand() % 1+1000;
int ok = 0;
while (ok == 0)
{
ok = 1;
for (int i = 0; i < *size; i++)
if ((*myStudents)[i].cod == code)
{
code = rand() % 1 + 1000;
ok = 0;
}
}*/
(*myStudents)[*size-1].cod = 7;
printf("Your code is: %d. Do not forget it! ", 7);
}

void insertStudent(Student **myStudents, int *size)
{
*myStudents = (Student*)realloc(myStudents, newSize*sizeof(Student));
// ^ <- look here
This is a pointer to a pointer to your students. realloc() expects the pointer to the data originally allocated, so you most certainly have to pass *myStudents here!
Also change the code to use a temporary variable. realloc() may return NULL on error, in which case the original memory is still allocated and you have to free() it.
For calculating the size, it's better to use the expression syntax of sizeof (newSize * sizeof **myStudents) because this prevents errors when later changing the type.
For sizes, you should always use size_t, as this is guaranteed to hold any possible size of an object (int is not ...)
Further side note: converting to and from void * is implicit in C and it's arguably better style not to write this cast explicitly.
All in all, the code should be written like this
void insertStudent(Student **myStudents, size_t *size)
{
size_t newSize = *size + 1;
Student *newStudents = realloc(*myStudents, newSize * sizeof *newStudents);
if (!newStudents)
{
free(*myStudents);
*myStudents = 0;
return; // check for this error in calling code
}
*myStudents = newStudents;
*size = newSize;
// [...]
}

realloc() needs pointer to the memory to be reallocated which is *myStudents and not myStudents.
Change
*myStudents = (Student*)realloc(myStudents, newSize*sizeof(Student));
to
*myStudents = (Student*)realloc(*myStudents, newSize*sizeof(Student));

You are reallocting to myStudents. It is not your intention and it is not possible.
From standard 7.22.3.5
void *realloc(void *ptr, size_t size);
Otherwise, if ptr does not match a pointer earlier returned by a
memory management function, or if the space has been deallocated by a
call to the free or realloc function, the behavior is undefined. If
memory for the new object cannot be allocated, the old object is not
deallocated and its value is unchanged.
Earlier you had undefined behavior, you didn't pass the originally allocated memory's address. Rather you passed a local variable. You had undefined behavior.
Student* t = realloc(*myStudents, newSize*sizeof(Student))
if(t){
*myStudents = t;
(*size)++;
}
else {
perror("realloc failed");
free(*myStudents);
exit(EXIT_FAILURE);
}
Also as you are increasing the memory you should do increase it if the call is successful. Then consistently access *size-1 throughout the code, it is easier to handle and go through.
The right way to do it shown above. In case realloc returns NULL you won't lose the reference to the already allocated memory. Also with this comes the advise to check the return value of realloc. Casting is redundant in this case - don't do it.
In the scanf you can simply write
scanf("%s",(*myStudents)[newSize-1].firstName);
otherwise you were passing char (*)[] it expects char*.

Related

Why does free() cause an error in my code when it's there, but everything runs well when it is not there?

My program takes an arbitrary number of words at runtime and stores them in a dynamically-sized array of words.
Currently, my program runs well, except when I use free() to free up the memory of the temporary double pointer temp. I am not quite sure why it does this, as I thought it would cause errors if I didn't use it.
int wordSize = 10, arrSize = 1, i = 0;
char **stringArr, **temp;
char *input;
stringArr = malloc(arrSize * sizeof(char *));
puts("Accepting input...");
for (;;) {
if (i >= arrSize) {
arrSize += 1;
temp = realloc(stringArr, arrSize * sizeof(char *));
if (temp != NULL) {
stringArr = temp;
free(temp); // This is the line that is giving me issues; removing it works
} else {
puts("Could not allocate more memory");
return 0;
}
}
stringArr[i] = malloc(sizeof(input));
input = malloc(wordSize * sizeof(char));
scanf("%10s", input);
if (strcmp(input, "END")) {
strcpy(stringArr[i], input);
i++;
} else
break;
}
free(stringArr);
At the bottom of my program I use free() without any issues. How come it works OK here but not earlier on in the program.
I feel I am missing something about how free() works.
Note: this is my first program implementing malloc() and realloc(), so I am only just getting used to how they work. If you know of a better way to accomplish what I am doing that, please feel free to describe.
The free(temp); line is causing an error (later on) because, in the preceding line, stringArr = temp;, you are assigning the address that is stored in the temp pointer to that in the stringArr pointer. Thus, when you free the memory pointed to by temp you also free the memory pointed to by stringArr, because it is the same memory block. Copying a pointer's value from one variable to another does not make a (separate) copy of the memory.
Omitting the free(temp); line is correct, because that memory is freed later on, in the free(stringArr); call.
You must not free the reallocated array when reallocation was successful. If you do that, the code will modify this freed block, which has undefined behavior and you will have further undefined behavior when you later try and reallocate or free this block.
Note also the following:
pre-allocating stringArr with a size of 1 is not necessary. Just initialize stringArr to 0 and arrSize to 0. realloc() can take a null pointer and will behave like malloc().
stringArr[i] = malloc(sizeof(input)); is incorrect: it will allocate a char array with a size of 4 or 8 depending on the size of a pointer on the target architecture, not 11 bytes as it should.
if the wordSize is the maximum length of a word, you should allocate one more byte for the null terminator. The 10 in %10s must match the value of wordSize, which is cumbersome because there is no easy way to pass this to scanf() as a variable.
you do not check the return value of scanf(), causing undefined behavior in case of premature end of file.
you have memory leaks: input is allocated for each iteration but never freed, freeing stringArr without freeing the strings pointed to by its elements makes them inaccessible.
It would be more efficient to use a local array to try and read the words with scanf() and only allocate the string and reallocate the array if successful.
Here is a modified version:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main() {
int arrSize = 0;
char **stringArr = NULL;
char input[11];
puts("Accepting input...");
while (scanf("%10s", input) == 1 && strcmp(input, "END") != 0) {
char **temp = realloc(stringArr, (arrSize + 1) * sizeof(*stringArr));
if (temp != NULL) {
stringArr = temp;
} else {
puts("Could not allocate more memory");
break;
}
stringArr[arrSize] = strdup(input);
if (stringArr[arrSize] == NULL) {
puts("Could not allocate more memory");
break;
}
arrSize++;
}
puts("Array contents:");
for (int i = 0; i < arrSize; i++) {
printf("%i: %s\n", i, stringArr[i]);
}
for (int i = 0; i < arrSize; i++) {
free(stringArr[i]);
}
free(stringArr);
return 0;
}

HEAP CORRUPTION DETECTED memory leak in C

I got and error which says
Debug assertation failed and heat corruption detected
like everything is working good in my program but I get that error. Where is the memory leak here? I have to free that memory in the main because my functions need to return pointers.
My code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int *dynamic_reader(unsigned int n) {
/*making new array and checking if allocation succeeded*/
int *mem;
mem = malloc(n * sizeof(int));
if (!mem) {
printf("Memory allocation failed\n");
exit(-1);
}
/*letting the user to input the values for the array*/
int i = 0;
while (i < n) {
scanf("\n%d", &mem[i]);
i++;
}
/*printing the array just to make sure everything is good*/
for (int j = 0; j < n; j++) {
printf("%d ", mem[j]);
}
return mem;
}
int *insert_into_array(int *arr, unsigned int num, int newval) {
/*making new bigger array*/
int *newarr = realloc(arr, (num + 1) * sizeof(int));
/*adding the integer to this new array */
newarr[num] = newval;
printf("\n");
/*printing to make sure everything is correct*/
for (int j = 0; j < num + 1; j++) {
printf("%d ", newarr[j]);
}
return newarr;
}
int main(void) {
/*In dynamic_reader function I need to make an array which size is given as a parameter*/
/*In this case I choosed 3*/
int *arr = dynamic_reader(3);
int num = 3;
/*In insert_into_array function I need to add one integer to this array I made in dynamic_reader*/
/*The parameters are the array, the number of elements in the array already done and the integer I want to add*/
int *c = insert_into_array(arr, num, 9);
/*I free the memory here because I need to return the pointers of these arrays in the function so it cant be done there*/
free(arr);
free(c);
}
You are double freeing your memory. Check the documentation for realloc. Realloc will either 1) expand the passed buffer or 2) will allocate a new buffer, copy the data, and free the original buffer. When you do:
free(arr);
free(c);
You are double freeing a value that was either once already freed by realloc or already freed by the first free(arr)
Additionally, you should check if realloc fails (returns NULL) and if so, handle the case appropriately.
First you malloc an array, which you return to your main function as arr. Then in another function, you realloc where arr is an argument to the realloc, but some other pointer stores the results. Depending on what happened in realloc, you've either got arr and newarr pointing to the same location, or newarr pointing to a valid location and arr pointing to an invalid location that has been freed. Either way, freeing both of them at the end is a problem.
No need to free(arr), that is taken care of when you realloc() it. The pointer returned by realloc() will either point to memory which includes the original memory, or free the old memory after copying its content to a new, larger chunk of memory.

C program crashes when accessing next element in array of struct

void struct_tokens(struct Words **w_count, int size)
{
*w_count = (struct Words*)calloc(size, sizeof(struct Words));
char *temp;
int n = 0;
int i = 0;
printf("Value 1: %s\n", (*w_count[0]).word);
printf("Value 2: %s\n", (*w_count[1]).word);
}
My struct looks like this:
struct Words
{
char word[20];
unsigned int count;
};
It crashes as soon as I access the next element in the array, so w_count[0] is accessed but it cannot access w_count[1] without crashing. So the Value 1 prints to console but Value 2 does not, why is this?
The calloc call is wrong - it would be
*w_count = calloc(size, sizeof(struct Words));
Accessing the instance of the structure would be
printf("Value 1: %s\n", (*w_count)[0].word);
[] has higher precedence than *. In earlier case you were having undefined behavior (which lead to the crash).
Also note that the memory allocated will be initialized with 0. As a result the char arrays will contain nothing other than \0 - it won't print anything when used in printf.
Few points:
void* to struct Words * conversion is done implicitly - you don't need to cast it.
Check the return value of calloc in case it returns NULL you should handle it appropriately.
Free the dynamically allocated memory when you are done working with it. (using free() ).
Code wise the correct code for calloc would be like this:-
*w_count = calloc(size, sizeof(struct Words));
if( *w_count == NULL ){
perror("calloc");
exit(EXIT_FAILURE);
}

How to send 2D char pointer to a function for memory allocation?

#include<stdio.h>
#include<stdlib.h>
#include<malloc.h>
void createDynamicArrayForChar(int dimension, char **ptr)
{
ptr = (char**)malloc(dimension*sizeof(char*));
for (int i = 0; i < dimension; i++)
{
ptr[i] = (char*)malloc(20 * sizeof(char));
ptr[i] = "value";
}
}
int main()
{
char **ptrArray;
createDynamicArrayForChar(5, ptrArray);
printf("%s", ptrArray[3]);
getchar(); getchar();
return 0;
}
It gives some errors when I try to compile this codes. How can I solve this problem? How to send 2D char pointer to a function in C?
Firstly, as per the present code, I see two issues.
You're passing ptrArray to the function and trying to allocate memory inside the function. Please be aware, C uses pass by value for function argument passing, so, if you want to allocate memory to ptrArray and expect that to be refeclted back to the caller, without returning, you'll be needing to pass a pointer to that `ptrArray.
in the code
ptr[i] = (char*)malloc(20 * sizeof(char));
ptr[i] = "value";
You're leaking memory. Once you've allocated memory using malloc(), you should use strcpy() to copy the data into the allocated memory.
That said, some advice:
Please see why not to cast the return value of malloc() and family in C.
sizeof(char) is guaranteed to be 1 in C. Using that as a multiplier is not required.
Always check the success of malloc() before using the returned pointer.
You probably need this (no error checking and not debugged code):
void createDynamicArrayForChar(int dimension, char ***ptr)
{
*ptr = (char**)malloc(dimension*sizeof(char*));
for (int i = 0; i < dimension; i++)
{
(*ptr)[i] = (char*)malloc(20 * sizeof(char));
strcpy((*ptr)[i],"value");
}
}
or
char **createDynamicArrayForChar(int dimension)
{
char **ptr = (char**)malloc(dimension*sizeof(char*));
for (int i = 0; i < dimension; i++)
{
ptr[i] = (char*)malloc(20 * sizeof(char));
strcpy(ptr[i],"value");
}
return ptr;
}
int main()
{
char **ptrArray;
ptrArray = createDynamicArrayForChar(5);
...
Read also Sourav Ghosh's answer.

is this a nice way to allocate memory for an array of arrays? (C)

Okay, imagine I have a char**, would this be the correct way to allocate memory?
I mean: allocate memory for the char** itself and then for each char*...
char** fraseUsuario = NULL;
int length = 100, i = 0;
fraseUsuario = (char **) malloc(sizeof (char*)); //Not pretty sure
for (i = 0; i < 3; i++) {
fraseUsuario[i] = (char *) malloc(length * sizeof (char));
if (fraseUsuario[i] == NULL) {
printf("error\n");
return -1;
}
gets(fraseUsuario[i]);
}
for (i = 0; i < 3; i++) {
printf("%s\n", fraseUsuario[i]);
free(fraseUsuario[i]);
}
And btw, how exactly does free() work? I mean, when I call it at the end, with the debugger it seems as if it does "nothing", if "Hello" is stored in the array, it will continue to be stored there after the free call... is that the normal behavior?
What do you mean allocate memory for the char ** itself? You allocate memory for a variable on the stack when you define it. The following statement defines (allocates memory) fraserUsuario and initializes it to NULL.
char **fraseUsuario = NULL;
I think what you probably meant is how to dynamically allocate an array of char **, i.e., pointer to a pointer to a character. Then you again dynamically allocate an array for each element of the previous allocated array. Do not use gets. It's deprecated and unsafe to use. Use fgets instead. Also, please don't cast the result of malloc. You don't get any benefit and you can run into error if you forget to include the header stdlib.h which contains its prototype. Here's how you do it.
char **fraseUsuario = NULL;
int max_string_len = 100 + 1; // maximum string length. +1 for null byte
int num_string = 3; // number of strings to read
int i, j;
fraseUsuario = malloc(num_string * sizeof *fraseUsuario);
if(fraseUsuario == NULL) { // check for NULL
// handle the case
printf("not enough memory\n");
return -1;
}
for(i = 0; i < num_string; i++) {
fraseUsuario[i] = malloc(max_string_len * sizeof(char));
if(fraseUsuario[i] == NULL) { // check for NULL
printf("not enough memory\n");
for(j = 0; j < i; j++)
free(fraseUsuario[j]); // free memory before returning
free(fraseUsuario); // free memory before returning
return -1;
}
if(fgets(fraserUsuario[i], max_string_len, stdin) == NULL) {
// reading string failed
*fraserUsuario[i] = '\0'; // empty string
}
}
for(i = 0; i < 3; i++) {
printf("%s\n", fraseUsuario[i]);
free(fraseUsuario[i]); // free memory allocated for strings
}
free(fraseUsuario); // free memory allocated for pointers to strings
fraseUsuario = NULL;
When you call free on a memory address which you got by a call to malloc, the memory block is returned to the free pool on the heap. This memory block can then later be reused by malloc. Once you free memory, you have given up your ownership of it. It no longer belongs to you and attempting to use it is illegal and will result in undefined behaviour and likely segfault.
You only allocate memory for one char* but use three.
To fix this do:
#define STR_MAXIMUM (3)
...
size_t length = 100, i = 0; /* No need to use a signed type.
size_t is meant as index and size type. */
char ** fraseUsuario = malloc(STR_MAXIMUM * sizeof(*fraseUsuario));
for (i = 0; i < STR_MAXIMUM; ++i)
{
fraseUsuario[i] = malloc(length * sizeof(*fraseUsuario));
...
Also add error checking to system calls.
Also^2: Do not use gets() as there is no way for the compiler or the machine to prevent the buffer passed in from overflowing. Use fgets() instead.
fgets(fraseUsuario[i], length, stdin);

Resources