Error on char array declaration in GCC 3.3.4

Question

int main () {

   char b[100];
   for (int i = 1; i <= 10; i++ )
       scanf ("%c%*c", b[i]);
}

but am getting the error 'Format arguemnt is not a pointer'

How can i declare an array to get all values form the user?

EDIT :

#include <cstdio>
#include <stdlib.h>

using namespace std;


int p[100], b, bc;
char bb[100];

int main () {


        printf("Enter Count : ");
        scanf ("%d", &bc);
        for (b = 1; b <= bc; b++ ) {
            printf("Enter a char and integer: ");
            scanf ("%c%*c %d", &bb[b-1], &p[b-1]);
            printf ("\n Your Entries =>  %c,  %d", bb[b-1], p[b-1]);
        }


    return 0;
}

This is my source code.

Your declaration is ok. Try `char b[100]; fgets(b, sizeof b, stdin);` ... and post real compilable (or very nearly compilable) code. — pmg, May 07 '11 at 14:01
Not quite sure why this being so heavily downvoted/voted-to-close... — Oliver Charlesworth, May 07 '11 at 14:05
After the edit: @coderex: you can't use `b` as `int` **and** `char[]` simultaneously. Change one of the names and alter the code accordingly. — pmg, May 07 '11 at 14:31
@pmg Sorry That was a miskte while copying from orginal source, i changed as bb. but not working :( — coderex, May 07 '11 at 14:37

score 3 · Answer 1 · answered May 07 '11 at 13:42

3

How about:

scanf("%c", &b[i]);

scanf() needs to know the address of the variable, so that it can modify it.

answered May 07 '11 at 13:42

Oliver Charlesworth

267,707
33
569
680

no there is no error now, but when i tried to print its not showing – coderex May 07 '11 at 13:48
I need "%c%*c" this because I want to read an integer after each char, other wise the newline entry will break the execution – coderex May 07 '11 at 13:49
1

@coderex: Yes, you may need to modify this for your exact requirements. My code snippet above was to demonstrate the need to provide an *address*. – Oliver Charlesworth May 07 '11 at 13:49
@coderex: `printf("%c\n", b[i]);`. – Oliver Charlesworth May 07 '11 at 13:53
yes, but nothing is printing :( – coderex May 07 '11 at 13:54
1

@coderex: I cannot help you unless you provide some *actual* code. – Oliver Charlesworth May 07 '11 at 13:57
I willupdate the latest code, its working in one section and not in another section – coderex May 07 '11 at 14:16
@Oli Charlesworth please check my EDIT – coderex May 07 '11 at 14:24

score 2 · Accepted Answer · answered May 07 '11 at 14:50

#include <cstdio>

Apparently, you're coding in C++

#include <stdlib.h>

Oh! Wait. It is C after all.
@coderex: make up your mind what language you're using

using namespace std;

Oh! It is C++. My answer does not consider the C++ specifics which I don't know.

int p[100], b, bc;
char bb[100];

If you can avoid using global variables, your code will be easier to deal with.

int main () {


        printf("Enter Count : ");
        scanf ("%d", &bc);
        for (b = 1; b <= bc; b++ ) {

The idiomatic way is for (b = 0; b < bc; b++). Using the idiomatic way, you won't need to subtract 1 inside the loop to access the array indexes.

            printf("Enter a char and integer: ");
            // scanf ("%c%*c %d", &bb[b-1], &p[b-1]);

scanf() is notoriously difficult to use correctly.
Anyway the "%d" conversion specifier already discards whitespace so you can remove the strange stuff; also there's an ENTER pending from the last scanf call. Using a space in the format string gets rid of it.

            scanf (" %c%d", &bb[b-1], &p[b-1]);
            printf ("\n Your Entries =>  %c,  %d", bb[b-1], p[b-1]);
        }


    return 0;
}

Have fun!

I recommend you stick with one of `C` or `C++` (or `C#`, or `D`, ...). Don't try to write a multi-language source file. LOL — pmg, May 07 '11 at 14:56

score 0 · Answer 3 · answered May 07 '11 at 14:49

0

Error was, its reading the "enter key" as newline char after an integer value enter.

to avoid this I used %*c scanf ("%c%*c %d%*c", &bb[b-1], &p[b-1]);

thank you everybody.

answered May 07 '11 at 14:49

coderex

27,225
45
116
170

See my answer ... but `scanf(" %c%d", &bb[b - 1], &p[b - 1]);` "works" and is easier on the eyes and brain. – pmg May 07 '11 at 14:59

Error on char array declaration in GCC 3.3.4

3 Answers3