i am noob. why wont this work?

Code:

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#include <iostream>
#include <cstdio>
using namespace std;
void disp(char *str[5]) {
    int choice=0;
    for(;;) {
        cout<<"Choose statement to read. (1-5) Other to exit."<<endl;
        cin>>choice;
        if((choice<1)||(choice>5)) break;
        cout<<"Statement "<<choice<<" is \""<<str[choice-1]<<"\"."<<endl;
        }
    return;
    }
int main() {
    char *str[5]={"","","","",""}, *input;
    cout<<"Please enter 5 strings to be stored."<<endl;
    for(int i=0;i<5;i++) {
            gets(str[i]); //I think the problem is here but i dunno what it is
            if(i<4) cout<<"Please enter next string."<<endl;
            }
    disp(str);
    cin.ignore();
    cin.ignore();
    return 0;
}

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Thanks alot

Hello,

I believe because str is a string literal. After initialization, any alteration to your string is undefined. You can either allocate memory, or create a 2-dimensional array.

As a string literal, I mean:

Code:

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char *ptr = "Hello";

---

You cannot alter that string. Though, the pointer can still be configured to point elsewhere, but modifying the strings contents after run-time is undefined.


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Quote:

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Originally Posted by Panopticon

i am noob. why wont this work?

---

You just answered your own question.



Besides, you should provide us a little more information than just a block of code if you want meaningful answers. Even though we're probably able to figure the code out given some time, it makes you look extremly lazy.

ok mr. sang drax i admit im lazy

the problem i think is in the line

Code:

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gets(str[i])

---

why cant i just make that pointer point to a string in the string table after the user inputs it?
the whole program compiles fine but i get illegal op when i input the string

i dont really understand overflow's explanation of literals
shouldnt it work anyway since when the user inputs a string, the string itself has a memory address which can be pointed to

thanks

Let me break it down.

Your two dimensional string works somewhat like this.

A pointer is a variable that contains the address of a variable.

Let's look closer:

Code:

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char *str[5];

---

You have a pointer. 5 strings that are currently "". You initialized them as string literals. You cannot change them unless you assign them to another variable that can be modified, or allocate a fresh block of memory to it.

Simply, your pointer does not have an address that is yours to use. It is a temporary variable called a constant.


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But cant the memory address be changed depending on the user input?

How would I achieve this then?
Do i initialise the pointers to variables rather than cnstants then?
thanks

Hello,

You could change str to "str[5][50]" if you don't want to worry about pointers. And also, you could use fgets() instead of gets(). This way you could use the appropriate function, and not worry about overflow.

In this task, it would be useless to allocate memory to your string because you don't know how much you are getting.


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ill get this one day =/
thanks