what can that do ?

Same thing. But use glob

they won't scan and add file with .h

why I can't change like that

target_link_libraries(main ${PROJECT_SOURCE_DIR}/libs/libglfw3.a ${PROJECT_SOURCE_DIR}/libs/libjsoncpp.a ${PROJECT_SOURCE_DIR}/libs/libjsoncpp.dll.a opengl32)

link_directories(${PROJECT_SOURCE_DIR}/libs) target_link_libraries(main libglfw3.a libjsoncpp.a libjsoncpp.dll.a opengl32)

int funct1(int a) { if(a) return funct1(--a)+a; else return 0; } int main() { int a=7; printf("%d",funct1(a)); return 0; }

can anyone explain how it's printing 21

#include <iostream> int main() { int a = 7; a = (--a) + a; printf("%d", a); }

#include <iostream> int main() { int a = 7; a = (--a) + a; printf("%d", a); }

The recursive isn't work

it does work

when --a

a become 6(both two)

you can have a try

a become 6(both two)

Ow, my bad, srry i miss the point

int funct1(int a) { if(a) return funct1(--a)+a; else return 0; } int main() { int a=7; printf("%d",funct1(a)); return 0; }

return fuct1 (--a) + a; a = 7; return funct1 (6) + 6; // 6 return funct1 (5) + 5 + 6; // 11 return funct1 (4) + 4 + 5 + 6; // 15 ... return funct1 (0) + 0 + 1 + 2 + 3 + 4 + 5 + 6; // 21 And recursion will be stopped because 0 will be converted to false.

6 + 5 + 4 + 3 + 2 + 1 + 0

return fuct1 (--a) + a; a = 7; return funct1 (6) + 6; // 6 return funct1 (5) + 5 + 6; // 11 return funct1 (4) + 4 + 5 + 6; // 15 ... return funct1 (0) + 0 + 1 + 2 + 3 + 4 + 5 + 6; // 21 And recursion will be stopped because 0 will be converted to false.

thanks

return fuct1 (--a) + a; a = 7; return funct1 (6) + 6; // 6 return funct1 (5) + 5 + 6; // 11 return funct1 (4) + 4 + 5 + 6; // 15 ... return funct1 (0) + 0 + 1 + 2 + 3 + 4 + 5 + 6; // 21 And recursion will be stopped because 0 will be converted to false.

When you type funct (--a) + a; compiler calls funct (--a) what decrements a and then adds new value of a

This sounds like System V ABI, it indeed mandates those registers must be preserved across function calls. What is actually the x86 standard you mean?

Intel documentation and programming guide for x86 development

Intel documentation and programming guide for x86 development

which calling convention does Intel define? The last time I checked the SDM it did not define any calling convention (apart from enclave operations). Maybe I should take another look at it.

Intel documentation and programming guide for x86 development

I haven't finished to read all of those. But I think the calling convention is defined per ABI.

Or is this the same with context switch that the OS is obliged to follow?

Hello, I am Hossein, a doctoral researcher at the University of Padova, Italy. With the aid of the University of Padova and NXP Semiconductor, we started a project similar to the Unreal Engine for programming micro-controllers in Italy. Are you interested in helping us with it?

How are you doing

How are you doing today

How are you doing today

Nice

Ok were are you located in

Intel documentation and programming guide for x86 development

Ok, you wrote: > it is entirely unrelated to calling conventions But what exactly is that? Those register saving and restoring rules are calling convention. It differs per ABI.

Ok, you wrote: > it is entirely unrelated to calling conventions But what exactly is that? Those register saving and restoring rules are calling convention. It differs per ABI.

I don't understand

I don't understand

Which part? Please be specific.

Abuja Nigeria

Abuja Nigeria

What?

I live in Abuja

why do we need two pointers here(*(uintptr_t*))?

Hi everyone. I need help with static variable that is written in C. The purpose of the function is to concatenate "abc" string twice to static variable store. On second foo() call I remove the value that was written in the previous call. I already found out that you cannot check whether the variable was initialized or not. I need some work around here, preferably with temporary string in init. Second static variables, global variables, and sentinel values are not allowed. Desired output: "abc\nabcabc\n". Current output "abc\nabc\n". void init(char **store) { (*store) = (char *)malloc(sizeof(char)); (*store)[0] = '\0'; } int foo() { static char *store; char *string; string = "abc"; init(&store); store = strcat(store, string); printf("%s\n", store); } int main() { foo(); foo(); }

Hi everyone. I need help with static variable that is written in C. The purpose of the function is to concatenate "abc" string twice to static variable store. On second foo() call I remove the value that was written in the previous call. I already found out that you cannot check whether the variable was initialized or not. I need some work around here, preferably with temporary string in init. Second static variables, global variables, and sentinel values are not allowed. Desired output: "abc\nabcabc\n". Current output "abc\nabc\n". void init(char **store) { (*store) = (char *)malloc(sizeof(char)); (*store)[0] = '\0'; } int foo() { static char *store; char *string; string = "abc"; init(&store); store = strcat(store, string); printf("%s\n", store); } int main() { foo(); foo(); }

Why would you do that?

Why would you do that?

Why would I do what?

Hi everyone. I need help with static variable that is written in C. The purpose of the function is to concatenate "abc" string twice to static variable store. On second foo() call I remove the value that was written in the previous call. I already found out that you cannot check whether the variable was initialized or not. I need some work around here, preferably with temporary string in init. Second static variables, global variables, and sentinel values are not allowed. Desired output: "abc\nabcabc\n". Current output "abc\nabc\n". void init(char **store) { (*store) = (char *)malloc(sizeof(char)); (*store)[0] = '\0'; } int foo() { static char *store; char *string; string = "abc"; init(&store); store = strcat(store, string); printf("%s\n", store); } int main() { foo(); foo(); }

There are several things go wrong with your code. 1) You only allocate sizeof(char) of memory with malloc(), but you use more. 2) You are leaking memory by overwriting the old pointer returned by the malloc for the second foo() call. 3) Your foo() is defined as a function that returns an int, but it does not return anything.

Practically, you should also free() anything you malloc()'ed.

Hi everyone. I need help with static variable that is written in C. The purpose of the function is to concatenate "abc" string twice to static variable store. On second foo() call I remove the value that was written in the previous call. I already found out that you cannot check whether the variable was initialized or not. I need some work around here, preferably with temporary string in init. Second static variables, global variables, and sentinel values are not allowed. Desired output: "abc\nabcabc\n". Current output "abc\nabc\n". void init(char **store) { (*store) = (char *)malloc(sizeof(char)); (*store)[0] = '\0'; } int foo() { static char *store; char *string; string = "abc"; init(&store); store = strcat(store, string); printf("%s\n", store); } int main() { foo(); foo(); }

You wrote: > I already found out that you cannot check whether the variable was initialized or not. For this specific situation, you can. Uninitialized static variable lives in .bss section. This area is initialized to zero when the program starts. So you can assume: static char *store is the same with static char *store = NULL;

There are several things go wrong with your code. 1) You only allocate sizeof(char) of memory with malloc(), but you use more. 2) You are leaking memory by overwriting the old pointer returned by the malloc for the second foo() call. 3) Your foo() is defined as a function that returns an int, but it does not return anything.

1) Yes, it is correct. My bad. 2) Should I free it before overwriting? 3) Yes, I should have defined it as void. The code that I sent is basically a small snippet from a big function. I need to fix only the logic that I described above.

You wrote: > I already found out that you cannot check whether the variable was initialized or not. For this specific situation, you can. Uninitialized static variable lives in .bss section. This area is initialized to zero when the program starts. So you can assume: static char *store is the same with static char *store = NULL;

I'll fix the code and come back to it.

I'll fix the code and come back to it.

Ok, you're good. For what it's worth, I fixed your code. I don't think there is a chance to call free() here. But let's assume memory leak is acceptable for this case. #include <stdio.h> #include <stdlib.h> #include <string.h> void init(char **pp) { if (*pp != NULL) { *pp = realloc(*pp, strlen(*pp) * 2 + 1); } else { *pp = malloc(4); (*pp)[0] = '\0'; } } void foo(void) { static char *store; init(&store); strcat(store, "abc"); printf("%s\n", store); } int main(void) { foo(); foo(); return 0; }

Ok, you're good. For what it's worth, I fixed your code. I don't think there is a chance to call free() here. But let's assume memory leak is acceptable for this case. #include <stdio.h> #include <stdlib.h> #include <string.h> void init(char **pp) { if (*pp != NULL) { *pp = realloc(*pp, strlen(*pp) * 2 + 1); } else { *pp = malloc(4); (*pp)[0] = '\0'; } } void foo(void) { static char *store; init(&store); strcat(store, "abc"); printf("%s\n", store); } int main(void) { foo(); foo(); return 0; }

Extra caveat: In real world application, we need to handle ENOMEM, so don't always assume that malloc(), realloc() and friends are always returning a valid pointer (as they may be returning NULL in case run out of memory).

Extra caveat: In real world application, we need to handle ENOMEM, so don't always assume that malloc(), realloc() and friends are always returning a valid pointer (as they may be returning NULL in case run out of memory).

Thanks for your effort and time :) I did not know that statics are initialized on programm start. And thanks for advice.

why do we need two pointers here(*(uintptr_t*))?

What is even the context of your question? Why do we need? I don't know, why would you write that?

ive already understood, ty

I want to program a serial terminal with a easy user interface for Windows. I never used C++ before. I was checking and Qt could be a good option for it. What do you think?

hello guys ,can anyone please help me using graphics.h in code block its kinda urgent

Hi i need help fast

How is the programme structure for creating a tree that has n high and in the right ir is written the numbers from 1 to the file number

And in the left the letters

Its form is a trangule

Example: n=3 1 A12 BC123

Hi. Is anybody tried to setup Boost.TestExplorer in VSCode?

warning: control reaches end of non-void function [-Wreturn-type] 18 | }

Can anyone help me

#include <iostream> using namespace std; void main (void) { static int N = 10; int counter, sum, number; cout << " Enter a Positive Integer smaller" << " or equal to 10 (the counter) "; cin >> counter; cout << endl; sum = 0; do { cout << " Enter a Positive Integer " ; cin >> number; cout << endl; sum = sum + number; counter ++; } while ( counter < N ); cout << " The sum of the digits you have entered is = " << sum << endl; if ( ( sum % 2 ) == 0 ) cout << " " << sum << " is divisible by 2 " << endl; else cout << " " << sum << " is not divisible by 2 " << endl; // note here no need of return anything // because the return type of ' main ' is ' void ' } // end of main

Basiclly i nedd to replace the do {  } while ( ) loop with a for loop

Basiclly i nedd to replace the do {  } while ( ) loop with a for loop

Just change your do while to a for loop like this: for(; counter < N; ++counter)

#include <iostream> using namespace std; void main (void) { static int N = 10; int counter, sum, number; cout << " Enter a Positive Integer smaller" << " or equal to 10 (the counter) "; cin >> counter; cout << endl; sum = 0; do { cout << " Enter a Positive Integer " ; cin >> number; cout << endl; sum = sum + number; counter ++; } while ( counter < N ); cout << " The sum of the digits you have entered is = " << sum << endl; if ( ( sum % 2 ) == 0 ) cout << " " << sum << " is divisible by 2 " << endl; else cout << " " << sum << " is not divisible by 2 " << endl; // note here no need of return anything // because the return type of ' main ' is ' void ' } // end of main

And the return type of main is not void but int. main should always return an int

#include <iostream> using namespace std; void main (void) { static int N = 10; int counter, sum, number; cout << " Enter a Positive Integer smaller" << " or equal to 10 (the counter) "; cin >> counter; cout << endl; sum = 0; do { cout << " Enter a Positive Integer " ; cin >> number; cout << endl; sum = sum + number; counter ++; } while ( counter < N ); cout << " The sum of the digits you have entered is = " << sum << endl; if ( ( sum % 2 ) == 0 ) cout << " " << sum << " is divisible by 2 " << endl; else cout << " " << sum << " is not divisible by 2 " << endl; // note here no need of return anything // because the return type of ' main ' is ' void ' } // end of main

using void as function parameter in C++ is redundant. func(void) is same as func() . But in C they're different. Also why not !(sum % 2) ?

using void as function parameter in C++ is redundant. func(void) is same as func() . But in C they're different. Also why not !(sum % 2) ?

whats the difference in C?

whats the difference in C?

func(); takes unspecified number of arguments. func(void); takes zero argument.

func(); takes unspecified number of arguments. func(void); takes zero argument.

then can i give any args in a func declared as func() ?

using void as function parameter in C++ is redundant. func(void) is same as func() . But in C they're different. Also why not !(sum % 2) ?

It means the same in C as well given that it occurs in the definition and not the declaration. So void func(void) {} is the same as void func(){} in C just like in C++.

Having said that, it is always a better idea to use void in C when you want to convey that a function has an empty parameter list.

then can i give any args in a func declared as func() ?

Yes

Hi guys :) I have to create accdb using ADO with c++ #import "C:\Program Files\Common Files\System\ado\msado15.dll" no_namespace rename("EOF", "adoEOF") #include <iostream> using namespace std; int main(){ CoInitialize(NULL); setlocale(LC_ALL, "Russian"); try { _bstr_t scon = "Provider=Microsoft.ACE.OLEDB.12.0;Data Source=C:\\DB\\CPP\\test.accdb"; _ConnectionPtr con("ADODB.Connection"); con.CreateInstance(__uuidof(Connection)); con->Open(scon, (_bstr_t)"", (_bstr_t)"", adConnectUnspecified); _bstr_t strCmd = "INSERT INTO Employees(EmployeeID, FirstName, LastName, HireDate, City, Country) VALUES(10, 'Mary', 'Williams', '15/4/1993 12:00:00', 'New York', 'USA')"; con->Execute(strCmd, NULL, adCmdText); con->Close(); } catch (_com_error& e) { cout << e.Description(); } ::CoUninitialize(); } Thats what I've done, but file test.accdb is not created. Works catch with mistake, that he cant find C:\DB\CPP\test.accdb Can someone help? 🥺

Basiclly i nedd to replace the do {  } while ( ) loop with a for loop

For loop need three things in its condition parenthesis i.e (Int i , i>=10,i++)

#include <iostream> using namespace std; int main () { static int N = 10; int counter, sum, number; cout << " Enter a Positive Integer smaller" << " or equal to 10 (the counter) "; cin >> counter; cout << endl; sum = 0; for(int i;i<10;i++) { cout << " Enter a Positive Integer " ; cin >> number; cout << endl; sum = sum + number; counter ++; } cout << " The sum of the digits you have entered is = " << sum << endl; if ( ( sum % 2 ) == 0 ) cout << " " << sum << " is divisible by 2 " << endl; else cout << " " << sum << " is not divisible by 2 " << endl; // note here no need of return anything return 0;// because the return type of ' main ' is ' void ' } // end of main

#include <iostream> using namespace std; void main (void) { static int N = 10; int counter, sum, number; cout << " Enter a Positive Integer smaller" << " or equal to 10 (the counter) "; cin >> counter; cout << endl; sum = 0; do { cout << " Enter a Positive Integer " ; cin >> number; cout << endl; sum = sum + number; counter ++; } while ( counter < N ); cout << " The sum of the digits you have entered is = " << sum << endl; if ( ( sum % 2 ) == 0 ) cout << " " << sum << " is divisible by 2 " << endl; else cout << " " << sum << " is not divisible by 2 " << endl; // note here no need of return anything // because the return type of ' main ' is ' void ' } // end of main

No need for static int now...

Can anyone help with this? #include <stdio.h> #include <stdlib.h> int main() { int sz = 1; scanf("%d",&sz); int arr[sz],i; for (i=0; i < sz ; i++) { arr[i] = rand() % 100; printf("%d:%d\n",i + 1,arr[i]); } return 0; } Write return functions a pointer to an element with an arbitrary index-dimensional array of 10 elements I am quite new to C please sry if my code looks bad

Hi friends

#include <iostream> using namespace std; int main () { static int N = 10; int counter, sum, number; cout << " Enter a Positive Integer smaller" << " or equal to 10 (the counter) "; cin >> counter; cout << endl; sum = 0; for(int i;i<10;i++) { cout << " Enter a Positive Integer " ; cin >> number; cout << endl; sum = sum + number; counter ++; } cout << " The sum of the digits you have entered is = " << sum << endl; if ( ( sum % 2 ) == 0 ) cout << " " << sum << " is divisible by 2 " << endl; else cout << " " << sum << " is not divisible by 2 " << endl; // note here no need of return anything return 0;// because the return type of ' main ' is ' void ' } // end of main

You have not assigned any value of i here actually replace i with counter

And the return type of main is not void but int. main should always return an int

Can we take the main function as void???

Can we take the main function as void???

The signature for main in C is int main(void) or int main(int argc, char* argv[]) You can also use char** argv instead of char* argv[]

#include <stdio.h> #include <string.h> int main() { freopen("oout.txt", "w",stdout); freopen("iin.txt", "r",stdin); int a[100],b,c,i,j,n,x,k,y,z=-10000; scanf("%d",&n); for(k=0;k<n;k++) { for(i=1;i<=3;i++) scanf("%d",&a[i]); // for(i=1;i<=3;i++) // printf("%d",a[i]); x=(a[1]+a[3])-(2*a[2]); // x=(a[1]+a[3]+1)-(2*(a[2]-1)); // printf("%d",x); // printf("%d",(a[1]+a[3])-(2*a[2])); if(x==0) { printf("%d\n",x); } else { if(x<0){ while(1){ // for(j=0;j<=1;j++) // { a[3]+=1; a[2]-=1; y=(a[1]+a[3])-(2*a[2]); // printf("%d\n",y); if(y>0){ y=((a[1]+a[3])-1)-(2*(a[2]+1)); printf("%d\n",y); break; } } // } } else{ while(1){ a[3]-=1; a[2]+=1; y=(a[1]+a[3])-(2*a[2]); // printf("%d\n",y); if(y<0){ y=((a[1]+a[3])+1)-(2*(a[2]-1)); printf("%d\n",y); break; } } } } } return 0; } why time limit excidded in todays div 2 contest

in codeforce

The signature for main in C is int main(void) or int main(int argc, char* argv[]) You can also use char** argv instead of char* argv[]

Got it