I was struggling to fix a code today, then I come across something similar to:
typedef struct {
int a;
int b;
int c;
int d;
char* word;
} mystruct;
int main(i
malloc the outer struct will only allocate 1 byte memory pointed by *word
since it is a 'char *' type. If you want to allocate more than 1 byte of memory pointed by word
, there are 2 options:
char word[50]
instead of `char *'word
as well. Remember to call free
twice as well in this case.Add a second malloc
for whatever length (N) you need for word
mystruct* structptr = malloc(sizeof(mystruct));
structptr->word = malloc(sizeof(char) * N);
You are allocating only memory for the structure itself. This includes the pointer to char, which is only 4 bytes on 32bit system, because it is part of the structure. It does NOT include memory for an unknown length of string, so if you want to have a string, you must manually allocate memory for that as well. If you are just copying a string, you can use strdup()
which allocates and copies the string. You must still free the memory yourself though.
mystruct* structptr = malloc(sizeof(mystruct));
structptr->word = malloc(mystringlength+1);
....
free(structptr->word);
free(structptr);
If you don't want to allocate memory for the string yourself, your only choice is to declare a fixed length array in your struct. Then it will be part of the structure, and sizeof(mystruct)
will include it. If this is applicable or not, depends on your design though.
Use word=malloc(128);
this will allocate 128 bytes to your varible word,
as you can read here you need to allocate the char *
separately :
mystruct* structptr = malloc(sizeof(mystruct));
structptr->word = malloc(sizeof(WhatSizeYouWant));
When you allocate memory for structptr
, the pointer word
in the struct
has no valid memory to point. So you either malloc
a piece of memory for word
, too, or make word
point to another character.