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F. Iván
Asked: 2020-10-16 07:02:52 +0800 CST 2020-10-16 07:02:52 +0800 CST

How to correctly manage memory when changing data?

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I am new to Cand the main differences with other languages ​​in which I am more advanced is the use of memory.

I am doing the following exercise:

Given the definition of T in the Th file, it is proposed to implement: T_mdfyTitle(): Modifies the data of type T received as a parameter, with a new name. Note: Correct memory management must be performed.

being T:

typedef struct {
    char* name;
    int code;    
} T;

For me in Java or C# it would be as simple as a setter, so I've implemented:

tError T_mdfyTitle(T* object, const char* newName){
    object->name=newName;
    return OK;
}

Now... when I reread:

Note: Correct memory management must be performed.

I think I'm not doing something right. Does my code correctly manage memory? And going further, when do I know that I have to manage memory?

c

1 Answers

  1. Best Answer

    When do I know I have to manage memory?

    As a broad answer to that broad question, if your code uses pointers you may have to manage memory.

    Coming from and you will be familiar that classes that handle strings are invariant in those languages. In other words, when you concatenate two strings in these languages, you do not modify one of the strings, but rather a new one is generated:

    String hola = "Hola".
String mensaje = hola.Append(" mundo!");

    The language does not have these mechanisms: it handles raw memory. To store a string of characters you must request memory to accommodate all the characters of that string plus an additional one (to mark the end of the string with the null character '\0'), so following the previous example:

            char *hola = "Hola";
/* 1 */ char *mensaje = malloc(12); // 'Hola' (4) + ' mundo!' (7) + '\0' (1)
/* 2 */ strcpy(mensaje, hola);      // Copiar 'Hola'
/* 3 */ strcat(mensaje, " mundo!"); // Concatenar ' mundo!'

// … usar cadena

/* 4 */ free(mensaje); // Liberar memoria

    Concatenation target variable has no memory allocated

    1. You have to allocate memory to it ( malloc).
    2. Once it has memory it has no content so you have to give it content ( strcpy).
    3. Once it has content we haven't concatenated anything yet so we concatenate ( strcat).
    4. Once we have used the memory requested with mallocwe must release it with free.

    As you can see, each string lives in a memory space pointed to by a pointer, so in your implementation:

    tError T_mdfyTitle(T* object, const char* newName){
    object->name=newName;
    return OK;
}

    You are copying the pointer (not the pointed content). This is not an error a priori if it is your intention, but if the original string (the one pointed to by the argument newName) disappears, the object->namewill be pointing to invalid memory. The correct memory management would be:

    1. Query the length of the string pointed to by newNamewith strlen.
    2. Reserve the necessary memory (counting the '\0') about object->namewith malloc.
    3. Copy the content pointed to by newNameinto the memory pointed to by object->namewith strcpyor memcpy.
    4. When an instance of is to be freed T, it must be freed freeover its pointer name.
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