Pointers: Am I finally grokking them??

Does this code that I cranked out give some indication that I'm getting it, or am I just dreaming in Technicolor?

Code:

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#include <stdio.h>

int main(int argc, char *argv[])
{
        int  myAge = 65;
        //int *ptr = &myAge;
       
        int *ptr;
        ptr = &myAge;
       
        printf("The pointer lives at: %p\n", (void *) &ptr);
        printf("The pointer contain the address of where my age is stored at, which is: %p", (void *)ptr);
        printf("\nI'm de-referencing a pointer to fetch my age which is: %d\n", *ptr);

  return 0;
}

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Any and all other advise would be most welcome!
--
Duke

**q=*ptr;

you can always do multiple pointers, whats stored in **q would be the address to *ptr then the address of myAge would be stored in *ptr.

You can initialize on same line also int *ptr=&myage;

Quote:

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Originally Posted by Sorinx

**q=*ptr;

you can always do multiple pointers, whats stored in **q would be the address to *ptr then the address of myAge would be stored in *ptr.

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Thanks! I'll have to give that a try ....

Quote:

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You can initialize on same line also int *ptr=&myage;

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You bet!! I have that commented out in my code. I had tried it out both ways ...

Thanks for the input!

Quote:

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Originally Posted by dukester

Code:

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int *ptr;
ptr = &myAge;
printf("The pointer lives at: %p\n", (void *) &ptr);

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Note that in C, pointer types do not need to be cast to other pointer types. So you may write the preferred form:

Code:

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printf("Here is a pointer: %p\n", &ptr);

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Also consider how pointers can be used in functions. Suppose you want a function that reads in two integers. We cannot return them, because in C we can only return a single value. We can write the function using pointers:

Code:

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#include <stdio.h>
#include <ctype.h>

// getdigits(x, y): read two digits from the stdin into x,y
// assume the digits are each one character long and do not
// perform any error-checking
void getdigits(int *x, int *y)
{
    int c;
    // Skip whitespace
    while ((c = getchar()) == ' ' || c == '\n' || c == '\t')
        ;
    // Convert first digit to int
    *x = c-'0';
    // Skip whitespace
    while ((c = getchar()) == ' ' || c == '\n' || c == '\t')
        ;
    // Convert second digit to int
    *y = c-'0';
}

// main: demonstrate the getdigits() function
int main()
{
    printf("Please type two digits separated by whitespace: ");
    int num1, num2;
    getdigits(&num1, &num2);
    printf("First digit: %d\nSecond digit: %d\n", num1, num2);
    return 0;
}

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Quote:

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Originally Posted by Sorinx

**q=*ptr;

you can always do multiple pointers, whats stored in **q would be the address to *ptr then the address of myAge would be stored in *ptr.

You can initialize on same line also int *ptr=&myage;

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That doesn't seem right. Pointers to pointers are set up so that the value is another pointer's location.
So if you were doing an assignment, you could do:
q = &p;
where q is a pointer to pointer and p is a pointer. If q can be dereferenced, you can instead do:
*q = p;


**q is the same as dereferencing p.

Quote:

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Originally Posted by c99tutorial

Note that in C, pointer types do not need to be cast to other pointer types. So you may write the preferred form:

Code:

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printf("Here is a pointer: %p\n", &ptr);

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I've tried that, and although it compiles, I get the same warning:

warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int **’ [-Wformat]

So casting to (void *) makes the warning disappear.

Quote:

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Also consider how pointers can be used in functions. Suppose you want a function that reads in two integers. We cannot return them, because in C we can only return a single value. We can write the function using pointers:

[snip]

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You bet!! Pass arguments by value and by reference ....
Not there yet, but encountered them in Oberon2 etc

This is a guess; but, I am thinking you are NOT using a C compiler; instead you are likely using a C++ compiler.
Edit: Reason for my guess is that in other places C++ requires casts where C compilers do it without casts.
Edit2: My guess appears to be wrong I even added the option -Wformat to my MinGW C++ Compiler and I can not dup your error.

Tim S.

Quote:

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Originally Posted by dukester

I've tried that, and although it compiles, I get the same warning:

warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int **’ [-Wformat]

So casting to (void *) makes the warning disappear.



You bet!! Pass arguments by value and by reference ....
Not there yet, but encountered them in Oberon2 etc

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Note: tested this code and it compiled with no warnings using MinGW C and C++ Compilers.

Code:

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#include <stdio.h>
#include <stdlib.h>

int main()
{
    /* int *ptr = NULL; Tried using two stars and still no warning with MinGW GCC or G++*/
    int **ptr = NULL;

    printf("Here is a pointer: %p\n", &ptr);
    return 0;
}

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The C++ Compile Command

Code:

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mingw32-g++.exe  -std=gnu++0x -Wextra -Wall  -g  -Winvalid-pch  -Iv:\SourceCode\OpenSourceCode\Apps\IDEs\CodeBlocks\Projects\testc -c v:\SourceCode\OpenSourceCode\Apps\IDEs\CodeBlocks\Projects\testc\main.cpp -o obj\Debug\main.o
mingw32-g++.exe  -o bin\Debug\testc.exe obj\Debug\main.o

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Code:

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mingw32-gcc.exe -Wextra -Wall  -g  -Winvalid-pch  -Iv:\SourceCode\OpenSourceCode\Apps\IDEs\CodeBlocks\Projects\testc -c v:\SourceCode\OpenSourceCode\Apps\IDEs\CodeBlocks\Projects\testc\main.c -o obj\Debug\main.o
mingw32-g++.exe  -o bin\Debug\testc.exe obj\Debug\main.o   
Output size is 28.36 KB
Process terminated with status 0 (0 minutes, 0 seconds)
0 errors, 0 warnings (0 minutes, 0 seconds)

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Quote:

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Originally Posted by c99tutorial

Note that in C, pointer types do not need to be cast to other pointer types

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That's not true. Any pointer type can be implicitly converted to a void pointer, and vice versa (i.e. without a "cast", aka type conversion).

That does not carry across to other pointer types. There is no implicit conversion, for example, between a pointer to char and a pointer to float.


There is also the complication that a lot of modern C compilers are actually C++ compilers, and implicit conversions to/from void pointers are not permitted in C++.

Quote:

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Originally Posted by dukester

I've tried that, and although it compiles, I get the same warning:

warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int **’ [-Wformat]

So casting to (void *) makes the warning disappear.

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That is a sign that your C compiler is actually a C++ compiler. Or (also possible, but less likely in practice) that your compiler is a bit over-zealous in checking that printf() format specifiers match the type of corresponding arguments.

It's an error to match an 'int *' with the %p specifier because 'int *' may have a different internal representation to 'void *'. Since printf's a variadic function, it performs no type conversion on its arguments (except for the default argument promotions) that match the ellipsis in its declaration, so an explicit cast is actually necessary in this situation.

edit: This applies to 'int **' as well.

Quote:

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Originally Posted by Barney McGrew

It's an error to match an 'int *' with the %p specifier because 'int *' may have a different internal representation to 'void *'.

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Whether it has a different representation or not does not matter; the language guarantees that casting from (void *) to (FOO*) as well as from (FOO*) to (void*) is implicitly allowed

Example:

Code:

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int int_sort_asc(const void *a, const void *b) {
    const int *x = a;  // no cast needed
    const int *y = b;  // no cast needed
    return *x - *y;
}

void array_sort(int n, int a[n]) {
    qsort(a, n, sizeof(a[0]), int_sort_asc);
}

void array_pprint(int n, int a[n]) {
    printf("Array starts at address %p\n", a);  // no cast needed
}

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Don't know!!! But ....

I can tell you that I use SciTE as my go-to editor! Here's the complete output:

Code:

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>gcc  -Os  -std=c99 -pedantic test-pointers3.c -o test-pointers3
test-pointers3.c: In function ‘main’:
test-pointers3.c:11:2: warning: ignoring return value of ‘scanf’, declared with attribute warn_unused_result [-Wunused-result]
>Exit code: 0

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after compiling:

Code:

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#include <stdio.h>

int main(int argc, char *argv[])
{
        int x;
        int *ptr;
       
        ptr = &x;
        printf("Gimme a number: ");
       
        (void) scanf("%d", ptr);    // or scanf("%d", &x);  same thing
        printf("You input: %d\n", *ptr);
        getchar();

  return 0;
}

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As you can see, SciTE is set up to use gcc for C code. I made sure to nuke all references to g++ by hacking my SciTEUser.properties file. So unless somethings going on "behind the scene" which I'm not aware of, then I'd say that it's gcc that barfing the warnings.

But..... I'm a C noob!! So what do I know?? lol

@c99tutorial: The point is that it isn't converted at all when it's passed to printf. If you pass an 'int *' to printf, then printf will receive an 'int *', because it has no reason to convert it to any other type. The difference is that in your example you provide explicit types in your argument list for int_sort_asc, whereas printf doesn't. Compare the following function declarations:

Code:

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int int_sort_asc(const void *a, const void *b);
int printf(const char *format, ...);

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Do you see the difference?

@dukester: You should check the return value to ensure that an item was successfully converted.

Quote:

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These functions return the number of input items successfully matched and assigned, which can be fewer than provided for, or even zero in the event of an early matching failure. The value EOF is returned if the end of input is reached before either the first successful conversion or a matching failure occurs. EOF is also returned if a read error occurs, in which case the error indicator for the stream (see ferror(3)) is set, and errno is set indicate the error.

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* scanf(3) - Linux manual page

Quote:

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Originally Posted by dukester

I've tried that, and although it compiles, I get the same warning:

warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int **’ [-Wformat]

So casting to (void *) makes the warning disappear.

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As you have shown in post #11, you are using the -pedantic option with gcc. Thus it follows strict ISO C rules.
The C standard says that the argument for "%p" shall be a pointer to void. The C standard doesn't guarantee that every pointer type is of the same size as a pointer to void. That's why you get the warning when you use -pedantic.
See also this answer on StackOverflow.

Bye, Andreas