How to write any program how to start

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I'm working on files concept in C++ in my project. First I have tried to read a complete line from the console(i.e. any data user types in one line ) using getline() and It was successfull. Now I want to read multiple lines(like paragraphs) from the console. I don't know how to read such big content from the console. If you know please help me.

I'm working on files concept in C++ in my project. First I have tried to read a complete line from the console(i.e. any data user types in one line ) using getline() and It was successfull. Now I want to read multiple lines(like paragraphs) from the console. I don't know how to read such big content from the console. If you know please help me.

you can use while or for cycle

It need lines limit before user enters lines for (int line = 0; line <= maxline + 1; line++) { // writing to file } But I don't want to tell the lines limit. Untill I press ESC key whatever I write on console need to be stored in the file.

Maybe using conio or ncurses stuff ?

You mean getchar() or getch() ?

Else, you gonna need tui stuff

Can you say what is tui please ?

It use conio on windows/dos or ncurses on *nix

Can you name those functions ?

It should be getc() but i never get into TUI programming, i just assume it

How to use VS or VS code

How to take input from the user on vectors with out declaring the size ?

/* When i=2 , j=0 what will be the Output? In my logic it will go switch(1) means case 1:() which is empty so it should go to default: (1+3)= 4 but on the output I am getting 3 why? Can you please explain it? */ #include <stdio.h> main() { int i, j, x = 0; for (i = 0; i < 5; ++i) for (j = 0; j < i; ++j) { switch (i + j - 1) { case -1: case 0: x += 1; break; case 1: case 2: case 3: x += 2; break; default: x += 3; break; } printf("%d\n", x); } printf("\nx = %d", x); } /* Output: 1 3 5 7 9 12 14 17 20 23 x = 23Press any key to continue . . . */

use break

where? after default?

if no case match then it goes to the default

yes but my qus is When i=2 , j=0 what will be the Output? In my logic it will go switch(1) means case 1:() which is empty so it should go to default: (1+3)= 4 but on the output I am getting 3 why? Can you please explain it?

case 1 does not use break so your case 2 start run it also does not uses break then case3 then your use break so it will not go to the default

/* When i=2 , j=0 what will be the Output? In my logic it will go switch(1) means case 1:() which is empty so it should go to default: (1+3)= 4 but on the output I am getting 3 why? Can you please explain it? */ #include <stdio.h> main() { int i, j, x = 0; for (i = 0; i < 5; ++i) for (j = 0; j < i; ++j) { switch (i + j - 1) { case -1: case 0: x += 1; break; case 1: case 2: case 3: x += 2; break; default: x += 3; break; } printf("%d\n", x); } printf("\nx = %d", x); } /* Output: 1 3 5 7 9 12 14 17 20 23 x = 23Press any key to continue . . . */

case 1 does not use break so your case 2 start run it also does not uses break then case3 then your use break so it will not go to the default

I will focus on for loops and variable status 1.Loop 1 x = 0,i = 0 Loop 2 i < j -> False 2.Loop 1 x = 0,i = 1 Loop 2 j = 0 case 0 x = 0 + 1 = 1 break j = 2 i < j -> False Print x = 1 3.Loop1 i = 2, x = 1 Loop 2 j = 0 case 1 case 2 case 3 Fall through to case 3, reason: No break statement between x += 2 => x = 1 + 2 break print x = 3 Hence the 3

case 1 does not use break so your case 2 start run it also does not uses break then case3 then your use break so it will not go to the default

/* When i=2 , j=0 what will be the Output? In my logic it will go switch(1) means case 1:() which is empty so it should go to default: (1+3)= 4 but on the output I am getting 3 why? Can you please explain it? */ #include <stdio.h> main() { int i, j, x = 0; for (i = 0; i < 5; ++i) for (j = 0; j < i; ++j) { switch (i + j - 1) { case -1: case 0: x += 1; break; case 1: case 2: case 3: x += 2; break; default: x += 3; break; } printf("%d\n", x); } printf("\nx = %d", x); } /* Output: 1 3 5 7 9 12 14 17 20 23 x = 23Press any key to continue . . . */

I have another confusion here I can see when i=4 , j=1 , switch (4+1-1)=4, ( case 4: ) which is not present on the code. So now it should go to default: (5+3)=8 But on the output I am getting 7 why?

maybe while loop

Please help me with a single line of code........ YouTube failed to help me in this

well it should go to the default try using a printf statement in the default

if you dont mind can you write step by step?

why not use the debugger to see the actual execution

I dont know how to use it.

I have another confusion here I can see when i=4 , j=1 , switch (4+1-1)=4, ( case 4: ) which is not present on the code. So now it should go to default: (5+3)=8 But on the output I am getting 7 why?

Why it does'nt work char *a = "abc"; char *b = "def"; a[0] = b[0]; Or *(a+0) = 'A'; what's wrong with pointer

What are you trying to do?

Why it does'nt work char *a = "abc"; char *b = "def"; a[0] = b[0]; Or *(a+0) = 'A'; what's wrong with pointer

https://pythontutor.com/ then try this

you'll have to decalre it like char string[] = "..."; so that it will be copied to a modifiable area of memory

That's simple. Earlier we got 3. 4. Loop 1 i=3,x=3 Loop 2 j = 0 case 2,(Since i + j - 1 is 3 + 0 - 1 = 2) Fall through to case 3 x = x + 2 => 3 + 2 = 5 break print x=5 Loop 2 j = 1 case 3: (Since i + j - 1 is 3 + 1 - 1 = 3) x += 2 => x = 5 + 2 = 7 break print x= 7 Hence the 7

Yes , i know that , is there any way? (For pointer)

why can't you declare it as an array?

Hello everyone how could I find the courses online from C/C++

Here could can find some C based system programming courses and projects. Www.csepracticals.com

no metter if you declare it char *, a string literal will still be const. But it's not a problem of the function, it's the variable that is declared wrongly

with GUI

You can use Qt as GUI framework. It can also provide you some audio stuff but I don't sure if it will be enough for your project