I'm trying to iterate an array which has 5 elements, each one with 1024 bytes. How can I iterate each byte of each element?
My code: Class.hh:
static char *sheets[5];
Class.cc:
#define SHEET_SIZE 1024
Class::sheets[0] = new char[SHEET_SIZE];
Because if I do:
for(int i = 0;i<5;i++)
{
cout << sheets[i] << endl;
}
Wouldn't this print like the whole element and iterate one sheet at a time?
One issue is that you don't know the length of each character array. Better to use std::string or std::vector<std::string>.
Here's some code to iterate over the matrix:
for (int row = 0; row < 5; ++row)
{
// If row is a C-string then we could use a nul terminator.
// Otherwise we'll assume a maximum length.
static const int MAXIMUM_COLUMNS = 1024;
for (int column = 0;
(sheets[row][column] != 0) || (column < MAXIMUM_COLUMNS);
++column)
{
Do_Something(sheets[row][column]);
// Or
std::cout << sheets[row][column];
}
std::cout << "\n";
}
std::vector<std::string>.
char * sheets[5];, which is a 2D character array (or it can be accessed like one). My answer shows how you can iterate through char * [5] as a 2d matrix. I have not overhauled the OP's design, nor have I changed the data structure. I'm going through the intent, which changed since your answer.*sheets[3]? Is it the size of a pointer? Or is it the length of C-Style string?The value of sheets[i] is a char pointer which will hopefully be terminated with a zero char otherwise it will segfault your application. If the zero char is at the beginning or in the middle it will not print the entire array so yes, it will fail your requirement.
If you want to iterate over every char of every array you have to do a little bit more:
for(int i = 0;i<5;i++)
{
cout << "Sheet " << i << endl;
for ( int j=0; j<SHEET_SIZE; ++j ) {
cout << sheets[i][j];
}
cout << endl;
}
I get a segmentation fault. Create a minimal reproducible examplesheets, except for sheets[0] are null. You may not indirect through null pointers. In order to iterate elements pointed by a pointer, you must first initialise the pointer to point to elements.Wouldn't this print like the whole element and iterate one sheet at a time?
It depends on what you have in a sheet. If they are only printable characters (ASCII code between 32 and 126) ending with a null character (0x00), you would get an expected output (1023 characters). But, if your sheet contains non-printable characters you won't get a "good" output; and the print will be cut once a null character (0x00) appears within a sheet.
You can have more control over each char in a sheet if you walk each sheet just as any other array. For example you could treat each sheet as an array of bytes, not just as a string, and dump its contents in hex:
sheets array, and, for each sheet,#include <iostream> // cout
#include <numeric> // iota
const size_t number_of_sheets{5};
const size_t sheet_size{5};
void print(char* sheet)
{
for (int i{0}; i < sheet_size; ++i)
{
auto to_string = [](auto c) -> char {
if (0 <= c and c <= 9) { return c + '0'; }
if (10 <= c and c <= 15) { return c - 10 + 'a'; }
};
auto bottom_nibble{(sheet[i] & 0x0f)};
auto top_nibble{((sheet[i] >> 4) & 0x0f)};
std::cout << (i ? "\'" : "") << to_string(top_nibble) << to_string(bottom_nibble);
}
}
int main() {
static char* sheets[number_of_sheets];
for (int i{0}; i < number_of_sheets; ++i)
{
sheets[i] = new char[sheet_size];
std::iota(sheets[i], sheets[i] + sheet_size, i * 0x11);
std::cout << "Sheet " << i << ": ";
print(sheets[i]);
std::cout << "\n";
}
}
// Outputs:
//
// Sheet 0: 00'01'02'03'04
// Sheet 1: 11'12'13'14'15
// Sheet 2: 22'23'24'25'26
// Sheet 3: 33'34'35'36'37
// Sheet 4: 44'45'46'47'48
sheet?