I used strtok to tokenize an array. I wanted to store the char pointer, that strtok returns, into an array.
char exp[] = {"1000 + 1000"};
char operands[50];
p = strtok(exp, " ");
Now I wanted to store the value of p (that is 1000) into the operands[i] array. I tried it like this:
memcpy(&operands[i], p, 49);
But it only copies a single integer.
I'm guessing you don't actually want to copy the string pointed to by p into the array of characters operands. Instead it seems to me that you want operands to be an array of pointers to char, i.e.
char *operands[50];
Then you can just do
operands[i] = p;
(Note: i have to be a valid index, in the range 0 <= i < 50)
However, the above is a C solution to a C problem. If you're programming in C++ you should probably use std::string and std::vector instead:
std::vector<std::string> operands;
...
operands.push_back(p);
Of course, if you're programming in C++ you should not use character arrays and strtok at all, but use the functionality in the C++ standard library for the tokenizing as well.
char exp[] = {"1000 + 1000"};
char operand[50];
char* p = strtok(exp, " ");
memcpy(&operand[0], p, 5);
After running above code, in the char array operand,
the first 5 are assigned value
operand[0] = '1' operand[1] = '0' operand[2] = '0' operand[3] = '0' operand[4] = 0 // end character of char array
others are not assigned value '?'
operand[i] = '?'
strcpy(operands, p)?std::stringis your friend.strtokdoes - put a NUL character in toexp[]after the first token, such that yourmemcpy(which has undefined behaviour because 49 is way larger thansizeof exp) copies "1000\0+ 1000\0" and about 40 other garbage characters. Listen to Bathsjeba - if you usestd::stringyou're less likely to screw up, and if you do you've a somewhat better chance to get the wrong string output instead of crashing your program.