any suggestion on how to strip characters from char array pass as pointer in C++. i must use memcpy function to copy.
void foo(char *test)
{
char a[1] = {0};
char b[1] = {0};
char c[1]= {0};
memcpy(&a,&test[0],1);
memcpy(&b,&test[1],1);
memcpy(&c,&test[2],1);
cout << a <<endl;
cout << b <<endl;
cout << c <<endl;
}
int main()
{
char uibuffer[4] = "ABC";
foo(uibuffer);
return 0;
}
the current output is:
ABC��
BC��
C��
desired output is:
A
B
C
void foo(char *test)
{
/* Please note, you need to add one extra byte here for a terminator */
char a[2] = {0};
char b[2] = {0};
char c[2]= {0};
memcpy(&a,&test[0],1);
memcpy(&b,&test[1],1);
memcpy(&c,&test[2],1);
cout << a <<endl;
cout << b <<endl;
cout << c <<endl;
}
int main()
{
char uibuffer[4] = "ABC";
foo(uibuffer);
return 0;
}
OR
Think about improving your code by getting rid of arrays and memory copying. Simple char a = buffer[x] will do the trick.
a. &a will be point to pointer to array. Wee need memcpy(a,&test[0]) or memcpy(&a[0],&test[0]) or memcpy(a,test), which is the same.memcpy(a,&test[0],1) will work, but so will memcpy(&a,&test[0],1)&a and a points to the same memory. I need to reread C specs :)since you haven't created \0 terminated strings do:
cout << a[0] <<endl;
cout << b[0] <<endl;
cout << c[0] <<endl;
Don't make a, b and c an array. Probably won't compile but to illustrate.
void foo(char *test)
{
char a = 0;
char b = 0;
char c = 0;
memcpy(&a,&test[0],1);
memcpy(&b,&test[1],1);
memcpy(&c,&test[2],1);
cout << a <<endl;
cout << b <<endl;
cout << c <<endl;
}
int main()
{
char uibuffer[4] = "ABC";
foo(uibuffer);
return 0;
}
Don't make them arrays:
char a = test[0];
char b = test[1];
char c = test[2];
You can still use memcpy, if you must, just as you are now.
