Char to Int - C++ - c++

Question

I know there are already answered questions about it, but I have read most of them and still couldn't solve my problem.
I have a program that will read notes, keep them in a list and give options to the user to delete, change, or select a specific note.
I'm using this struct:
struct List {
char title [101];
char text [501];
int cont; //code of the note.
struct List* next;
};typedef List list;
I'm stuck at the point of selection, if the user types a * it must return all notes, and if the user type a number it must return only the corresponding note.
So far i just have this:
List* select (List *l, int v) {
List *p = l;
for (p = l; p != NULL; p = p -> next){
if( p -> cont == v){
cout << "\nTitle: " << p -> title << "\n";
cout << "Text: " << p -> text << "\n";
cout << "Code: " << p -> cont << "\n" << "\n";
}
}
how can I read a symbol in char and transform to int, to compare it with the code of the note.
sorry if i write something wrong, i'm brazilian and i have none practice for writing.
EDIT:
Thank you so much guys, it realy realy helped me a lot and now i get to finish my work! :D

Answer 1

Try this:
Get the input as a string
string str;
cin >> str;
If the string is a *
if (str == "*")
{
// print out all notes
}
Else, attempt too convert string into number with strtol. Strtol gives better error checking than atol because it tells you whether the whole string was converted and if not, where so you can go look.
char * endp;
long code;
code = strtol(str.c_str(), // the string converted to characters
&endp, // where in the string the number ended.
10); // the base of the number, base 10 in this case for
// decimal. 16 if you want to input in hex.
// Unlikely in this case.
if (*endp == '\0') // If the number didn't end didn't end at the end
// of the string, the number is invalid
{
// print note for code
}
A quick note on strtol(str.c_str(), &endp, 10); and testing endp.
This doesn't actually work. endp winds up pointing at a memory location that may not be valid by the time you get to check. Really, you need to get the char array out of the string and into something for which you can guarantee the scope.
The above warning is incorrect based on old, or faulty data. Thank you #TamásSzabó. Makes future code writing a wee bit simpler.
No std::string version is almost the same:
char str[64]; // allocate storage for a big number.
cin.get(str, sizeof(str)); // read up to size of string -1 and null terminate
if ((str[0] == '*') && (str[1] == '\0'))
// first character in string is '*' and there
// is no second character
{
// print out all notes
}
else
{
char * endp;
long code;
code = strtol(str,
&endp,
10);
if (*endp == '\0') //this time endp will be valid
{
// print note for code
}
}

Answer 2

If you really need to only char and int types, you could try this:
char buf[10]; // Assuming that you won't have more than 10 characters in your number
char *endp;
cin >> buf; // This may cause you problems if the input string is more than 9 characters long!
From here on you can use user4581301's answer:
if ( (buf[0] == '*') && (buf[1] == '\0') )
{
// Print all your notes here
}
code = strtol(but, // the string converted to characters
&endp, // where in the string the number ended.
10); // the base of the number, base 10 in this case for
// decimal. 16 if you want to input in hex.
// Unlikely in this case.
if (*endp == '\0') // If the number didn't end didn't end at the end
// of the string, the number is invalid
{
// print note for code
}