why *ptr is not working while using float array, did I mistake with the wrong format specifier or so-CodePudding

In this code i was trying to print the array using pointer but it just doesn't work with float array. Note: i have done this exact same code with int array and it was working fine and as expected but this on the other hand is miserable. Float array is the problem i guess.

CODE:

#include<stdio.h>

int main(){

    

    float arr[]= {1, 2, 3, 4, 54};
    int *ptr ;
    ptr = arr;


    printf("%u\n", ptr);
    printf("%u\n", *ptr);
    printf("%d", &ptr);

}

i am expecting this output.

OUTPUT: address of arr[0] . value of arr[0] . address of ptr

CodePudding user response：

To begin with you use a pointer to an int to point to your array. That needs to be float * to be correct:

float *ptr = arr;

Then lets take the printf statements one by one...

We start with:

printf("%u\n", ptr);

Here ptr is a pointer. The format %u is to print an unsigned int value. To print a pointer you need to use the %p format. And the pointer need to be a void * pointer. Mismatching format specifier and argument type leads to undefined behavior.

Second:

printf("%u\n", *ptr);

Once you fix the pointer issue mentioned first, then you use *ptr which is the same as ptr[0] which is a single float value. The recommended format to print a float (or a double) value is %g or %lg. Again you have mismatching format specifier and argument type.

Lastly:

printf("%d", &ptr);

Here &ptr is a pointer to the variable ptr itself. It will have the type float **. As before you have mismatching format specifier and argument type.

To solve your problems you need to use correct specifier and arguments:

printf("%p\n", (void *) ptr);
printf("%g\n", *ptr);  // ptr[0] would also work
printf("%p", (void *) &ptr);

Most compilers available today should be able to detect these kind of problems and issue warnings about them. You might need to enable more warning to get the warnings. Always treat warnings as errors that must be fixed.

CodePudding user response：

int *ptr ; tells the compiler that ptr is a pointer to an int. Then ptr = arr; attempts to assign a pointer to a float to this pointer to an int.

The compiler prints a warning or error message when you do that. Do not ignore warnings from your compiler. Enable warnings in your compiler and elevate warnings to errors. With Clang, start with -Wmost -Werror. With GCC, start with -Wall -Werror. With MSVC, start with /W3 /WX.

Change int *ptr ; to float *ptr;.

printf("%u\n", ptr); is not a correct way to print a pointer. Use %p and convert the pointer to void *: printf("%p\n", (void *) ptr);. Do the same for &ptr: printf("%p\n", (void *) &ptr);. Also, do not leave \n off when printing; always print a new-line character at the end of the line. (This is useful both for buffering issues and for making the end of your program’s output clear, not mingled with your command-line shell prompt.)

printf("%u\n", *ptr); is not an correct way to print a float. After changing ptr to float *, use printf("%g\n", *ptr);.