copy_to_user: treating more data

copy_to_user: treating more data

By : user2951333
Date : November 18 2020, 01:01 AM
To fix this issue In my kernel module I have the following read function: , Firstly, in your code for sample_read() change
code :
copy_to_user( buffer, &c, 1 );
copy_to_user( buffer, &c, sizeof(c));
typedef struct data {
    int x;
    int c;
} data_t;

data_t val;
val.x = gpio_get_value(BTN_2);
val.c = gpio_get_value(BTN);

copy_to_user( buffer, &val, sizeof(data_t));

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How can i process the data after '/ ' in url with out treating it as directory?

How can i process the data after '/ ' in url with out treating it as directory?

By : attief
Date : March 29 2020, 07:55 AM
this one helps. You need to use a .HTACCESS file with RewriteEngine On.
create a file in the root directory named .htaccess (just .htaccess, nothing else!) - if it already exists, add this to the bottom.
code :
<IfModule mod_rewrite.c>
    RewriteEngine On
    RewriteRule ^([A-Za-z0-9]+) /index.php?params=$1 [L]
copy_to_user() and copy_from_user() for basic data type

copy_to_user() and copy_from_user() for basic data type

By : user3418393
Date : March 29 2020, 07:55 AM
should help you out If the function receives a pointer to user-space data, you have to use copy_from_user() to copy the pointed-to data from user space into kernel space (and vice versa).
Note that the pointer value itself is passed by value (like all C parameters), so you don't have to do a copy_from_user() to obtain the pointer value before you can copy_from_user() the data it points to.
copy_to_user not copying data?

copy_to_user not copying data?

By : Suvarna Kodukula
Date : March 29 2020, 07:55 AM
hope this fix your issue Are you sure about return 0;? I think this function should return the amount of bytes copied, in your case this should be return sizeof(read_value32);
copy_to_user does not return expected data

copy_to_user does not return expected data

By : user2477020
Date : March 29 2020, 07:55 AM
wish helps you Your parens are off on the call to copy_to_user. I think you mean to test if the response of copy_to_user is != 0. Instead, you're putting sizeof(ABC_T) != 0 as the last argument. Since sizeof(ABC_T) is non-zero, your call ends up being copy_to_user((void __user *)arg, &abc, true).
Fix your parens and see if you get better results:
code :
if (copy_to_user((void __user *)arg, &abc, sizeof(ABC_T)) != 0) {
    // ...
Why can I print this treating as a reference and treating it as a scalar?

Why can I print this treating as a reference and treating it as a scalar?

By : HajarRoot
Date : March 29 2020, 07:55 AM
seems to work fine The $a2d[0] is an array reference. We can take this array reference and print out the 3rd entry:
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