convert Linked list in C++ to C

Hellow everyone,

I have a problem in converting a C++ code to C.

Code:

---

struct node
{
int id;
string name;
int idP;
string nameP;
double price;
int unit;
node *nextC; //idk how to convert to c
};


node *temp; //idk how to convert to c
node *headC = NULL; //idk how to convert to c

---

then, there are a part that using this.. which is :

Code:

---

temp = new node;
temp-> id= tempid;
temp->name= tempname;
temp-> idP= tempidP;
temp->nameP= tempnameP;
temp-> price= tempprice;
temp->unit= tempunit;
temp-> nextC = headC;
headC= temp;

---

please help :( I totally no idea how to convert them.. TT

Code:

---

struct node
{
int id;
string name;
int idP;
string nameP;
double price;
int unit;
struct node *nextC;
};


struct node *temp;
struct node *headC = NULL;

---

Look up the "malloc" function to replace "new"; remember "free" function will likely be needed.

Tim S.

Quote:

---

Originally Posted by stahta01

Code:

---

struct node
{
int id;
string name;
int idP;
string nameP;
double price;
int unit;
struct node *nextC;
};


struct node *temp;
struct node *headC = NULL;

---

Look up the "malloc" function to replace "new"; remember "free" function will likely be needed.

Tim S.

---

Hellow, thanks for your fast respond, really appreciate it. :)

I tried :

Code:

---

temp = malloc(sizeof(node));
temp-> id= tempid;
temp->name= tempname;
temp-> idP= tempidP;
temp->nameP= tempnameP;
temp-> price= tempprice;
temp->unit= tempunit;
temp-> nextC = headC;
headC= temp;


---

and it throw an error

Code:

---

Undeclared identifier 'node'.

---

><

EDIT : Nvm, I found that I should use temp = malloc(sizeof(struct node)); instead :)

Quote:

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Originally Posted by Milo Juak MuTou

Nvm, I found that I should use temp = malloc(sizeof(struct node)); instead :)

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It is usually easier to use the pattern

Code:

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temp = malloc(sizeof *temp);

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as you don't actually need to remember the type of temp this way. sizeof is an operator that only looks at the size of the operand, here the data type pointed to by temp.

Just remember to read that line of code as "dynamically allocate one temp".

If you wanted to allocate memory for an array of say five elements, then

Code:

---

temp = malloc(5 * sizeof *temp);

---