I have to compare 2 strings, one is from a structure member and the other is "Empty".
I use strcmp like so:
if (strcmp(e[n]->seat[seat/5][(seat%5)-1], "Empty")==0)
I am getting the error Invalid conversion from char to const char*.
Can someone please help me out with this?
It sounds like e is an array of structures each containing a 3 deminsional character array. You should probably be using a two deminsional char pointer array unless you /know/ the string length won't change but if you don't just use the address like chnossos said.
if (strcmp(&(e[n]->seat[seat/5][(seat%5)-1]), "Empty")==0)
Related
I'm trying to create an array of pointers using my professors example in class. I can't seem to get it to work though.
char * ISBN[] = {
"1-214-02031-3",
"0-070-21604-5",
"2-14-241242-4",
"2-120-12311-x",
"0-534-95207-x",
"2-034-00312-2",
"1-013-10201-2",
"2-142-1223",
"3-001-0000a-4",
};
This is my professors example of declaring an array of pointers of type char. He says this is what we should do for the assignment. Sadly, I'm getting the error -
Error (active) E0144 a value of type "const char *" cannot be used to initialize an entity of type "char *" -
I've asked about this before and someone said that char * has been deprecated. But the professor says that this is the declaration that we should use for the assignment. How would I go about trying to figure out how to get this array working? What exactly is this declaration doing?
const char * ISBN[] = {
"1-214-02031-3",
"0-070-21604-5",
"2-14-241242-4",
"2-120-12311-x",
"0-534-95207-x",
"2-034-00312-2",
"1-013-10201-2",
"2-142-1223",
"3-001-0000a-4",
};
I guess I just needed to add const in front of char *. I still got much to learn. Thanks everyone.
It is simply because you are trying to assign string literals to char pointer.In c++ string literals are const char pointer because as string literals are stored in read-only memory.Another reason is for optimisation for compilers(storing only one instance of a literal that is repeated many times in the source). So you get multiple pointers to the same memory, instead of each occupying a separate chunk of memory.
I initialized a char and a pointer, and then point it to the char. But I found that only the first character in the char was passed into the pointer. How could this be? Appreciate any advice!
char p[]="This is a long char.";
std::cout<<sizeof(p)<<"\n";
char *ptr=p;
std::cout<<*ptr<<"\n";
Got:
21
T
Program ended with exit code: 0
When checking the value of variables I noticed that:
ptr=(char *) "This is a long char."
*ptr=(char) 'T'
Why is the ptr holding the whole char but the content in it is only the first character? Isn't this the opposite of what we call a pointer? Got really confused...
When you said:
std::cout<<*ptr...
you dereferenced ptr, which gives you the (first) character that ptr is pointing to.
flyingCode's answer shows you how to print the entire string.
In C++, an array is just like multiple variables grouped together, so they each have a memory address, but they are one after another.
So, the pointer is pointing at the address of the first element, and the other elements are in the adresses right after that in order.
When you use the * operator on a pointer, it references the contents of the memory address the pointer is pointing to.
If you wanted to access another element of the array using this, you would use *(ptr + 5) to offset the memory address.
When you use the variable name alone, the language just does the offset for you and gets all the contents of the array. The pointer doesn't know the length of the array, but it can find the end of it, because char arrays always end with a null character (\0).
ran your code in repl and changing it to the code below worked :D
char p[]="This is a long char.";
std::cout<<sizeof(p)<<"\n";
char *ptr=p;
std::cout<<ptr<<"\n"; //notice no *
I recommend you use std::string rather than c-strings unless you have a specific need for them :)
I tried to use a character Pointer go throw string character (iterate) but I found i can not say the below:
string Name = "Hello";
char *ch = Name;
like the previous statements i am getting error during execution.
However when I am doing like that:
char *ch = "Hello";
the program running without throwing any exception.
Why is that?
I have recently encountered similar problem and the simplest answer is that std::string is a different type from char*, more precisely std::string is an object which contains some characters (your text) and few methods, which allow you to do multiple operations with your text. You can imagine creating a class Integer for storing the value, but also a method allowing you to square and cube the number which is stored in the Ingerer class. Even though they could store the same numerical value, you will not be able to compare them (unless you overload the operator==), as their types are different.
If you wish to use the code you provided, you need to rewrite the second line as
const char *ch = Name.c_str();
it is allowed because std::string contains a method c_str() which "casts" itself to const char*. If you want to learn more about strings, be sure to visit C++ reference about strings.
What is the difference between
char (CharBuff[50])[10];
and
char CharBuff[10][50];
My requirement is to have 10 character buffers, each of length 50 max (including null terminating character)
And, I would like to access the buffers as CharBuff[0], CharBuff[1] and so on.
char (CharBuff[50])[10];
declares CharBuff as a 50-element array of 10-element arrays of char. The parentheses are superfluous in this case.
char CharBuff[10][50];
declares CharBuff as a 10-element array of 50-element arrays of char.
Given that you want 10 strings of up to 50 characters, you would use the second form; the type of each CharBuff[i] will be "50-element array of char".
If you really wanted to create a separate type definition for a 50-element array of char, you could do something like
typedef char Str[50];
...
Str CharBuff[10];
Now CharBuff is a 10-element array of Str, which is a 50-element array of char.
Normally, I would not create a separate typedef like this unless I wanted to make Str opaque; that is, I don't want to expose the details of its implementation to whomever's using it. In addition to the typedef, I'd also supply an API for allocating, assigning, copying, formatting, and displaying objects of type Str.
Put another way, if the person using the Str type has to be aware that it's a 50-element array of char in order to use it properly, then it's better to just make them use a 50-element array of char.
The other answers here which say they're the same are wrong! Both are NOT the same arrays and their sizes are very different. Here's a snippet to show that:
char i[50][10];
std::cout << sizeof(i[1]) << '\n';
char (j[10])[50];
std::cout << sizeof(j[1]) << '\n';
10
50
You can see the live example here.
i is a 50-element array with each element being a 10-element array of characters, while j is a 10-element array with each element being a 50-element array of characters. Although the total sizes of both would be the same, the size of an element at each level would be different. If you assume they're the same, it would lead to undefined behaviour
i[25][5] // OK
j[25][5] // accessing j beyond index 9 is undefined behaviour!
This shows that the parenthesis have no significance in a non-pointer, non-reference array declaration i.e. char (j[10])[50] is just confusing notation for char j[10][50].
My requirement is to have 10 character buffers, each of length 50 max
Then you should declare your array as char CharBuff[10][50].
Nobody ever uses the former, always use the latter form:
char CharBuff[10][50];
Latter syntax will be proficient and less confusing as per you requirement of 10 rows (char buffers) having length of 50 each.
Go for the last one as it clearly states what you are allocating.
However, since the question is tagged with c++ too, I would advise using std::vector and std::string as it follows:
std::vector<std::string> CharBuff(10, std::string(50, '\0'));
The big problem that i see here is confusing syntax. When you use parenthesis they already have known roles in c and c++ syntax such as type conversions, pointers to functions(which looks a bit like what you wrote). What you are using them for add's a new meaning which makes code more obfuscated and ignores the fact that c and c++ have a great and intuitive way, for anybody that used matrices at least once, to express 2d arrays. So, for a clean and not confusing syntax use the latter version:
char CharBuff[10][50];
You are looking for 10 arrays of 50 chars each.
To declare a single buffer of 50 bytes, you would use
char CharBuff[50];
But you want to have 10 buffers, so just tack that on to before[50], e.g.:
char CharBuff[10][50];
Now CharBuff[0] will address the first fifty-byte buffer, CharBuff[1] will get the second, and so on.
so I'm implementing this chess program on C++ and I'm trying to integrate to winboard protocol...one of the functions that they say I need to write to do so should have the following signature:
char *MoveToText(MOVE move); // converts the move from your internal format to text like e2e2, e1g1, a7a8q.
my question is....the text formats are something like e2e2....but the return type of that function is char...and as far as I can understand it, char is just one single character....
so how come are they telling me to use this signature?
or am I mistaken and in fact char can also store multiple characters such as e2e2, e1g1 etc?
Yeah, in C, a char* points to an array of characters. C treats arrays of characters as strings, terminated by a null byte.
The return is a char* or a c-style string =)
char * is a pointer on char - address of sequence of characters.
It returns pointer to char, which is basically a c-string.
Take a look at this tutorial: http://www.cprogramming.com/tutorial/lesson9.html