Help me ...

Hello@

i have written this

Code:

---

#include <stdio.h>


struct account_info
{
        char name;
        char surname;
        int  account;
        int  rest;
} a;


main()
{
        printf("Give name:");
        scanf("%c\n", &a.name);



        printf("You gave %c\n", a.name);
}

---

i put John, but in the end the programme print Only the 1st letter, J.

what is the problme?

THXXXXXXXXXXXXXX

'name' is only a single character. You will need an array, and you'll need to use something other than %c to read input. Read the FAQ for how to get input from a user.


Quzah.

Quote:

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Originally Posted by quzah

'name' is only a single character. You will need an array, and you'll need to use something other than %c to read input. Read the FAQ for how to get input from a user.


Quzah.

---

you have right..

I change it char -> char* and then i put %s...

but now the exit is

you gave (null)

Quote:

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Originally Posted by johnybe

you have right..

I change it char -> char* and then i put %s...

but now the exit is

you gave (null)

---

try defining your name variable as

Code:

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char name[15];

---

that defines an array of chars 15 bytes long. plenty of space for a first name

Quote:

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Originally Posted by ollie88r

try defining your name variable as

Code:

---

char name[15];

---

that defines an array of chars 15 bytes long. plenty of space for a first name

---

Well, you had right!!! Thxxx!!!

But now i have problem with this. If i want to do the same thing not for the name, but forthe account

Code:

---


#include <stdio.h>

typedef struct account_info
{
        char* name[15];
        char* surname[15];
        int  account;
        float  rest;
} a;


main()
{
        a x;
        printf("Give account:");
        scanf("%i", x.account);



        printf("You gave %i \n", x.account);
}

---

i have "segmentation fault" when i try to put 12234.

????

scanf takes an address, not a value.

Code:

---

scanf("%i", &x.account);

---

thx you!!!!!

:)

thxxxxxxxxxxxxxxxxxxxxxx

Code:

---

{
        char* name[15];

---

You have an array of pointers to characters. You don't want that. You want to remove the *, or remove the [15] and allocate memory dynamically. Pick one, not both.


Quzah.

Quote:

---

Originally Posted by quzah

Code:

---

{
        char* name[15];

---

You have an array of pointers to characters. You don't want that. You want to remove the *, or remove the [15] and allocate memory dynamically. Pick one, not both.


Quzah.

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ok...!! thx you tooo!!!!!!!!!!!!