In math.h you will only get definition of the functions you are using not the whole source code

/usr/include/math.h Is a header file You have to check the gnu libc code repository on their website or github because you only get the binaries

thanks alot

This group chat doesn't allow me to send a link here You can easily find it by simply googling "glibc source code"

/usr/include/math.h Is a header file You have to check the gnu libc code repository on their website or github because you only get the binaries

There are tons of libc, you should check the libc source code that you used

Hi to all. I tried to do as I was advised but still nothing works, refuses to accept input at all (35 lines of code). If I just use scanf, it only remembers what goes before the interrupt, and I need to enter a two-word book title. Maybe someone knows what I'm doing wrong? I will be grateful for the advice. https://onlinegdb.com/XMoGNEShIk

use int main not void main

But I use int main

You also has no NULL handling if malloc fails, and using printf rather than putchar() to output single char

You also has no NULL handling if malloc fails, and using printf rather than putchar() to output single char

I'm sorry, but I don't really understand why the symbol, if I need the word itself?

I'm sorry, but I don't really understand why the symbol, if I need the word itself?

I've told you several times, If you input something with scanf using enter, there will be remained newline on input stream

I make something like scanf ("[^\n] %c", ptr[i]. name, & instr); It would be better or You not mean that?

I haven't try it, but my practice is scanf(); getchar()

I did so, but now the problem is that the cycle has become endless

Oh sorry I miss one think, now cycle is Okey

Although everything is fine with the cycle now, but even so it does not want to enter the title of the book

I use scanf(%[^\n] %s, ptr[i]. name, & instr) ; getchar () It is like I nead or I'm wrong again?

how do i create a linked list data with struct?

first line will input into ptr[i].name because the specifier will read the input until found newline

EXAMPLE 1 repeat.txt 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 output 1 0 1 0 0 1 0 1   EXAMPLE 2 repeat.txt 16 1 2 49 189 16 1 2 49 5 2 16 1 2 49 189 16 1 2 49 5 2 16 1 2 49 189 16 1 2 49 5 2 16 1 2 49 189 16 1 2 49 5 2 16 1 2 49 189 16 1 2 49 5 2 16 1 2 49 189 16 1 2 49 5 2 output 16 1 2 49 189 16 1 2 49 5 2  

Hello. Who can explain why I assign one value to variable, but in this variable I got another value, after assigning? This what I got in GDB: 66 size_t l_idx = strchr(seq, tolower(letter)) - seq; (gdb) n 67 (gdb) p l_idx $1 = 17 (gdb) p strchr(seq, tolower(letter)) - seq $2 = 469123456210287 (gdb) wtf

Suppose say that the letter is not found and strchr returns a null pointer. Then subtracting seq from it results in Undefined Behavior. The difference in the outputs could be because you are converting a ptrdiff_t to size_t in the former case and printing the ptrdiff_t in the latter case. In any case, doing such things may result in Undefined Behavior. The right way to do this would be to check if the return value from strchr is not a null pointer and only then subtract seq from it

Thx anyway, really weird > the right way to do this would be to check if the return value from strchr is not a null pointer Sure, no way. There is an “if” statement above, and strchr in this case physically can’t return NULL. Here it is: … if (!((letter >= ‘A’ && letter <= ‘Z’) || (letter >= ‘a’ && letter <= ‘z’))){ fprintf(stderr, “error: \’%c\’: not a letter. Abort.\n", letter); exit(EXIT_FAILURE); } const char *seq = “qwertyuiopasdfghjklzxcvbnm”; size_t l_idx = strchr(seq, tolower(letter)) - seq; … Buut just in case I also have tried same, but now “letter” is definitely in string, ‘a’ for example. And I tried it without “tolower()” function. Same results.

You can use isalpha() to get rid of the ugly comparison inside the if statement also it is usually more optimised and faster than simply comparing things

but this might not be the reason you are getting the bug

how do i create a linked list data with struct?

how do i create a linked list data with struct?

Thx anyway, really weird > the right way to do this would be to check if the return value from strchr is not a null pointer Sure, no way. There is an “if” statement above, and strchr in this case physically can’t return NULL. Here it is: … if (!((letter >= ‘A’ && letter <= ‘Z’) || (letter >= ‘a’ && letter <= ‘z’))){ fprintf(stderr, “error: \’%c\’: not a letter. Abort.\n", letter); exit(EXIT_FAILURE); } const char *seq = “qwertyuiopasdfghjklzxcvbnm”; size_t l_idx = strchr(seq, tolower(letter)) - seq; … Buut just in case I also have tried same, but now “letter” is definitely in string, ‘a’ for example. And I tried it without “tolower()” function. Same results.

To this exact struct in this struct

Why are you subtracting "seq" from the letter

Why are you subtracting "seq" from the letter

i want do with this struct: struct node { int data; struct node *next; }

i want do with this struct: struct node { int data; struct node *next; }

1. Allocate first struct 2. Allocate other structs within this struct by: node = node->next; node->next = malloc(sizeof(struct node)); 3. Fill it by using the same way of iterating list as in ‘2’ 4. And then print values in list using the same way of iterating as in ‘2’

Thx anyway, really weird > the right way to do this would be to check if the return value from strchr is not a null pointer Sure, no way. There is an “if” statement above, and strchr in this case physically can’t return NULL. Here it is: … if (!((letter >= ‘A’ && letter <= ‘Z’) || (letter >= ‘a’ && letter <= ‘z’))){ fprintf(stderr, “error: \’%c\’: not a letter. Abort.\n", letter); exit(EXIT_FAILURE); } const char *seq = “qwertyuiopasdfghjklzxcvbnm”; size_t l_idx = strchr(seq, tolower(letter)) - seq; … Buut just in case I also have tried same, but now “letter” is definitely in string, ‘a’ for example. And I tried it without “tolower()” function. Same results.

Make sure you sure you passed the right type to tolower() And also check if you have got any compiler warnings those often warn you if you handled types incorrectly

Can anyone help me explaining the output of this C program. The output is like this: 7 7 12 49 https://pastebin.com/w2VcHXp1

i want to input some integers untill the given number is -1 and save them a a linked list! how can do it?

i want to input some integers untill the given number is -1 and save them a a linked list! how can do it?

do { // read num // push it to the linked list } while (num != -1);

What if the first element is itself num = -1

do { // read num // push it to the linked list } while (num != -1);

I assume you to know how to create and append elements into linkedlist. Let's say your input is n; then... while(n != -1) { // add into linked_list }

how can i push it to the linked list?

you gotta google that since you may be using your own implementation of a linked list or maybe some lib for that...

i think you need to use at least a struct

yes of course struct node { int data; struct node *next; } but this question must be easier than things i found in google because the purpose of the question is to become more familiar with Pointer and struct, not the concepts of objective programming

#define PRODUCT(x)(x*x) PRODUCT(i+1) is replaced by i + 1 * i + 1, and i = 3 so thats 3 + 3 + 1 = 7 note that x*x should be (x)*(x) if you want the right result (16) PRODUCT (i++) = i++*i++, which is 3*4, since the ++ operator is computed after the *, and now i was incrimentented twice (i = 5) PRODUCT(++i) = ++i*++i = 7*7, since the increment is executed before the multiplication

#define PRODUCT(x)(x*x) PRODUCT(i+1) is replaced by i + 1 * i + 1, and i = 3 so thats 3 + 3 + 1 = 7 note that x*x should be (x)*(x) if you want the right result (16) PRODUCT (i++) = i++*i++, which is 3*4, since the ++ operator is computed after the *, and now i was incrimentented twice (i = 5) PRODUCT(++i) = ++i*++i = 7*7, since the increment is executed before the multiplication

Oh thank you so much You are right. I forgot that if you want to use that kind of statement then you have to put them inside braces.

Your code is wrong irrespective of whether you put the parentheses around x or not. Your code exhibits Undefined Behavior. Read the link in the message above this or refer to the Resources section of this channel.

I read it. From the looks of it, I believe that it is happening because I am trying to modify the variable i more than once between two sequence points. By the way, will I be able to circumvent this problem by performing the incrementation in separate lines twice before the computation?