c++ - incompatible pointer type warning in c? -

uint32 measurements [32];  xcp_addr_t xcpapp_convertaddress( uint32 address, uint8 extension ) {      return &measurements[address];  //return incompatible pointer type   } 

address value recieving client (example : 0,1,2.....). above function return address of measurement other internal function. getting warning below :

return incompatible pointer type 

could tell me how solve ?

it depends on type of xcp_addr_t. expression is: &measurements[address]. type of expression convertible uint32*. if return type uint32 in disguise, remove & operator. if return type different, rethink doing.

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so typedef is:

typedef uint8* xcp_addr_t; 

as can see uint8* (of return type) , uint32* (the returned value's type) don't match. can either change return type or type of array measurements to:

uint8 measurements[32]; 

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ok, want ensure xcpapp_convertaddress returns valid pointer (without going out of bounds). have 2 choices:

• assert it
• throw exception

you can assert doing:

assert(address < 32); return &measurements[address]; 

in case program fail @ runtime if address passed function incorrect (notice have add #include <cassert> use assert).

alternatively can throw exception:

if (address < 32) throw std::runtime_error("out of bounds"); return &measurements[address]; 

(notice you'll need #include <stdexcept> std::runtime_error).