/* Answer to Q/A: Why "int m=('o'<<8)+'y'; cout<<(char*)&m;" outputs "yo" https://www.sololearn.com/Discuss/837932/sum-of-char-in-a-int */ #include #include using namespace std; int main() { // ASCII: 'y' = 0x79; 'o' = 0x6f // http://www.asciichars.com/_site_media/ascii/ascii-chars-landscape.jpg // Don't mind the complicated output, I just wanted everything to be in one format // Declare an integer, initialize it to 0 int m = 0; cout << "Zero integer:" << endl; cout << setfill('0') << setw(8) << hex << m << endl; // Shift the value of ASCII character 'o' (6f) by 8 to the left, assign it to m // ('o'<<8) in original code m = 'o' << 8; cout << "Integer containing ascii character 'o' (6f) shifted by eight to the left:" << endl; cout << setfill('0') << setw(8) << hex << m << endl; // Declare another integer, assign a value of ASCII char 'y' (79) to it int n = 'y'; cout << "Integer containing ascii character 'y' (79):" << endl; cout << setfill('0') << setw(8) << hex << n << endl; // Add them together // +'y' in original code m = m + n; cout << "Adding them togeter:" << endl; cout << setfill('0') << setw(8) << hex << m << endl; // Create C-string (char*), point it to our integer. // (char*)&m in original code char *c_string = (char*)&m; // As x86 platform is little endian, the least significant byte is stored at the lowest address // so our integer will be stored like this (from smallest memory address to greatest): 796f0000 // which gives us 'y' 'o' '\0' '\0' in ASCII encoding. // As C-strings are null terminated, every byte will get printed out until we reach a zero value // so 'y' 'o' gets printed cout << "Our C-string:" << endl; cout << c_string; return 0; }

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