无法从char转换为void *

Question

SAFEARRAYBOUND bound[1] = {25, 0};
SAFEARRAY * psa = SafeArrayCreate(VT_UI1, 1, bound);
for(long int i = 0; i <25; i++)
    SafeArrayPutElement(psa, &i,sendBuf[i]);

I am trying to creating a safearray from a char array but get this error cannot convert from char to void* sendBuf is a char array

Answer 1

You didn't show us the error, but it appears SafeArrayPutElement takes a pointer to the element as the third parameter. I believe it'll work if you use:

SafeArrayPutElement(psa, &i, &sendBuf[i]);

Note the & .

Answer 2

It would be more efficient in this case to use SafeArrayAccessData() instead of SafeArrayPutElement() :

SAFEARRAYBOUND bound[1] = {25, 0}; 
SAFEARRAY * psa = SafeArrayCreate(VT_UI1, 1, bound); 
void *pvData;
SafeArrayAccessData(psa, &pvData);
memcpy(pvData, sendBuf, 25);
SafeArrayUnaccessData(psa);

Or:

SAFEARRAYBOUND bound[1] = {25, 0}; 
SAFEARRAY * psa = SafeArrayCreate(VT_UI1, 1, bound); 
unsigned char *pvData;
SafeArrayAccessData(psa, (void**)&pvData);
for(long int i = 0; i <25; i++)     
    pvData[i] = sendBuf[i];
SafeArrayUnaccessData(psa);