If you are struggling with with syntax it needs just time and practice

Ziky do you mind if I message you in DM, I don't want to intrude or go against group rule please

Sir I am affraid you have only one life :D

Actually my struggle is not syntax in particular, my issue is developing things, bringing all the knowledge together, developing working programs that's my issues

U can try to solve some simples Codewars problems, just to get the feeling on how to solve problems using C https://www.codewars.com/ After you being capable of easily solve fundamental problems you can try a simple project from some "C projects list" on the internet, i will provide you a link if you wanna see it https://www.codewithc.com/c-projects-with-source-code/

Please I do want the project list thanks

It's in the last link in the upper message

Сan you help me with tht task please 1. Create a text file representing a fragment of the C++ program text. There is one operator in each line of text. Replace each assignment statement of the form "variable = variable + number;" to the abbreviated assignment operator of the form "variable += number;". Determine the number of changed operators and write this value as the "number" value in the last changed operator of the assignment of the input text. Display the contents of the input and modified files.

Ulyana Andreieva sent a code, it has been re-uploaded as a file

If I have an input line like a=a+1; your logic would fail as you determine pos to be at position of '=' and then search for '+' from pos+3 which is out of bounds. Your idea seems to be right. I would suggest reading the line, find just the variable name and the input number which can be found by using rfind for +. All the code you have now can be used for error handling, like suppose if a line is not in the format expected. But be careful with the bounds. Once you determine these 2, write the line <var name>+=<num> to output file. And just keep count of how many such replacements are being made.

🏔️ sent a huge message, it has been re-uploaded as a file hello, do you know what could be the problem in this code? like "memory override" and other stuf..

🏔️ sent a huge message, it has been re-uploaded as a file hello, do you know what could be the problem in this code? like "memory override" and other stuf..

🏔️ sent a huge message, it has been re-uploaded as a file hello, do you know what could be the problem in this code? like "memory override" and other stuf..

i don't know but it was in my exam 🥲

If this question was in your exam and the expected answer is not Undefined Behavior then the person who asked you this question is an idiot

Is it not implementation defined behaviour while writing in the struct and unspecified behaviour afterwards?

Guys know how to create the if statement that counts the letters from the file txt?

It is Undefined Behavior. He is copying the string "what could go wrong" into an array of length 5. This is what the C standard says wrt to your question on writing into the strict: An array subscript is out of range, even if an object is apparently accessible with the given subscript (as in the lvalue expression a[1][7] given the declaration int a[4][5]) (6.5.6). So even if we were to assume that he was writing only a string of length 9 into a and this overflows into b, it is still Undefined Behavior.

🙃🙃Hello everyone, I have a question regarding memory allocation. Well when we declare a variable and associate it to a data type we are reserving a space in memory so far so good. Depending on the architecture, the storage size can vary, for example, type “int” in 32bits, its size is 2bytes and 4bytes in 64bits. If I have architecture for 64bits int value = 132255; Will the variable have allocated four addresses where each address will store 4 bytes? Or In just one address will it store the entire value? The other question associated with this is the concept of data type is it related to a set of numbers available in the type? or in the amount of bytes it can store and when stored it breaks into parts according to the amount of bytes? In this case, when 132255 is stored, it will be placed in an address or divided into four because the int type for 64 bits has 4 bytes.

First question, it will be one address, it will point to the start of the memory that was allocated, and since the size of the type is known to the compiler it can handle reading from that memory correctly when you dereference the pointer. The second question I don't really understand, are you talking about C or C++? because concept of data types differs between the languages

First question, it will be one address, it will point to the start of the memory that was allocated, and since the size of the type is known to the compiler it can handle reading from that memory correctly when you dereference the pointer. The second question I don't really understand, are you talking about C or C++? because concept of data types differs between the languages

🙃🙃Hello everyone, I have a question regarding memory allocation. Well when we declare a variable and associate it to a data type we are reserving a space in memory so far so good. Depending on the architecture, the storage size can vary, for example, type “int” in 32bits, its size is 2bytes and 4bytes in 64bits. If I have architecture for 64bits int value = 132255; Will the variable have allocated four addresses where each address will store 4 bytes? Or In just one address will it store the entire value? The other question associated with this is the concept of data type is it related to a set of numbers available in the type? or in the amount of bytes it can store and when stored it breaks into parts according to the amount of bytes? In this case, when 132255 is stored, it will be placed in an address or divided into four because the int type for 64 bits has 4 bytes.

Hi i am trying to understand when should i use the restrict keyword

C++

The thing is, no array subscript is actually used here, only pointers to characters

LDR R0, [R0,#PluginContext.field_14] LDR R0, [R0, #4] LDR R0, [R0] LDRB R3, [R0, #7]

LDR R0, [R0,#PluginContext.field_14] LDR R0, [R0, #4] LDR R0, [R0] LDRB R3, [R0, #7]

Struct layout in the case of consecutive arrays of char requires no padding bits between the arrays. Two consecutive char arrays will be stored consecutively. So a and b in the struct will be stored consecutively here. Moreover for a struct of char arrays no sane implementation will allocate padding bits before or after the arrays (this is what is implementation defined behavior according to you). On all implementations, the size of this struct would be 9. This is not guaranteed by the standard but this is what it will be. Now writing beyond allocated memory irrespective of whether you do it via a pointer or through the array directly is Undefined Behavior and not Unspecified Behavior.

Every data type has known to a compiler size, if you compile to a platform where int is 4 bytes, then the compiler knows that int is always 4 bytes, so it knows how much data to allocate and how to address it. In general for a variable of any type to be used correctly: 1. Memory for it should be allocated 2. Constructor should be called on that memory. When you call new it does both for you. If you create a variable on the stack, then both will also be resolved automatically (memory on the stack will be prepared for your variable and constructor called when needed). Even if the constructor doesn't do anything (and there's no actual call happens under the hood) formally this rule is true.

But this is allocated memory, as it is the stack itself, I do not know how unregulated access of the stack is defined in the standard

int n; int **arr= new int *[n] What does it means

int n; int **arr= new int *[n] What does it means

Guys when you declare a pointer this way Int arr[20] If you do this in a function it allocates in the heap or in the stack

Hi im learning C whats this declaration mean/do? hexstr = (char *) malloc(size*BYTE_MULT); I know that * Is used for pointer declaration but only (Char *) what do? Thanks for any help

Hexstr = Is not a declaration?

It's not a declaration, it's a type cast

No, it's an assignment

Hi im learning C whats this declaration mean/do? hexstr = (char *) malloc(size*BYTE_MULT); I know that * Is used for pointer declaration but only (Char *) what do? Thanks for any help

This line of code is using the malloc() function to dynamically allocate memory in C. In this particular case, hexstr is being assigned a pointer to the memory block that malloc() allocates. The (char *) is a type cast which tells the compiler to treat the pointer returned by malloc() as a char pointer. This is because malloc() returns a pointer of type void *, which can be cast to any other pointer type. size and BYTE_MULT are variables that are being multiplied together to determine the size of the memory block to allocate. Once the memory has been allocated, hexstr can be used as a pointer to an array of char values that has a size of size * BYTE_MULT. It's worth noting that when you allocate memory using malloc(), you are responsible for freeing it when you're done using it. This is typically done using the free() function.

We need ChatGPT here..