Question

I am trying to allocate an array of structs within a function.

My struct is as follows:

 typedef struct{
      uint16_t  taskNumber;
      uint16_t  taskType;          
      double    lat;               
      double    lon;               
      double    speed;             
      uint8_t   successCriteria;   
      uint16_t  successValue;      
      uint8_t   nextPoint;      
 }missionPoint;

In my code I declare a missionPoint-pointer which I then pass into the function that will dynamically allocate it after parsing a file and figuring out how big it needs to be.
Currently this is how my code looks:

 missionPoint* mission;    //declaring the pointer
 parseMission(mission);

The parseMission function will then parse a specific file and find out how many missionPoints I need and will then allocate it in the following manner:

 mission = (missionPoint*) malloc(n * sizeof(missionPoint));

where n is the parsed number of missionPoints I need.

The problem is that within the function I can see the proper values but not outside of it; once the function returns it’s like nothing happened.

I would appreciate your help in making it so that the function modifies the original pointer and I can see the data from outside the function.

Answer 1

You need to pass a reference to the pointer, i.e. a double pointer, because the address itself is going to be modified:

missionPoint *mission;
parseMission(&mission);

The argument of parseMission should now be of type missionPoint ** instead of missionPoint *:

*mission = (missionPoint*) malloc(n * sizeof(missionPoint));

This wouldn’t be necessary if you only wanted to modify the memory mission is pointing to, but it cannot be avoided since you are assigning a new value to the pointer itself.

Also note that casting the return value of malloc is not necessary in C.