|
David Crow wrote: You're showing your age I hope that I have a few more years left in this worn out body.
David Crow wrote: Looking at ordinal numbers inside of Windows DLLs is going back aways. Those were the good ol' days.
There isn't a more accurate way to describe that particular DLL export. It's exported from shell32 with the NONAME attribute.
D:\Tools>dumpbin /exports "C:\Windows\System32\shell32.dll"
...
61 00102C60 [NONAME]
... In fact it looks like there are 355 additional exports in shell32 that are NONAME (only exported by ordinal)
Best Wishes,
-David Delaune
|
|
|
|
|
What are you actually trying to do? Do you just want to run this dialog, or start some other process?
|
|
|
|
|
Richard, all I want to do is just what windows 10 does, but access from the sidebar, instead of
windows + R. Is there a programming way of doing it?? Thanks.
|
|
|
|
|
Access the Run dialog from the sidebar? What is the sidebar?
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
|
|
|
|
|
Run the command line:
rundll32 shell32.dll,#61
|
|
|
|
|
|
Problem c++
Hi i have an txt-file that looks like this:
Bigman arnold :line1
1333131313131 :line2
blue 123 green 2, 234 green :line3
arctic 34 :line4
The thing i have to do is modify the out put from the txt file.
There are many more lines.
I have to modify it so the output from the files so it becomes like this:
arnold Bigman :line1
blue 123 green 2, 234 green :line 2
arctic 34 :line3
1333131313131 :line4
And the lines continues for the rest of the txt file.
I'we been trying ifstream with a while(file>>string1>>string2>>getline(zzzzz, file)
But i get an error im totally of.
Sometimes there are only 3 lines for the adress and sometimes 4.
Any good ideas?
|
|
|
|
|
So are you needing to know how to read from one file and write to another, how to swap strings, or something else? I see no pattern in your input/output (e.g., line 1 swaps the two words, line 2 moves to line 4).
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
|
|
|
|
|
begzz wrote: But i get an error We cannot guess what error you get. Please edit your question, show the actual code you are using, the error that you get and where in the code it occurs.
|
|
|
|
|
PS C:\Users\anubi\Desktop> cd "c:\Users\anubi\Desktop\" ; if ($?) { g++ Tryout1.cpp -o Tryout1 }
; if ($?) { .\Tryout1 }
c:/mingw/bin/../lib/gcc/mingw32/9.2.0/../../../../mingw32/bin/ld.exe: cannot open output file Tryout1.exe: Permission denied
collect2.exe: error: ld returne
I can't even run a simple program to print "hello". I installed MinGW and setup the path. What is mising in VSC?
|
|
|
|
|
The message is clear. Either you do not have permission to write into that location, or the file is in use by some other process.
|
|
|
|
|
The most common cause of that error is the executable your trying to compile is already running, probably from a previous launch of the app you never closed.
|
|
|
|
|
In C, I need to code the next 10 prime numbers of n entered from the keyboard with an array. How do I do it?
|
|
|
|
|
|
iHuy wrote: How do I do it? Can you do it on paper? If not, then trying to code such an algorithm is not going to work (unless someone just outright gives you the code).
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
|
|
|
|
|
I know Borland turbo IDE had a compiler that supported multiple multiple inheritance in C++. I want to know is there any other compiler that does that or was that it?
|
|
|
|
|
All C++ compilers support multiple inheritance. That includes Visual Studio, g++.
This is a language feature that hasn't been deprecated yet. If it is a good idea to use it, that's another question
Mircea
|
|
|
|
|
All C++ compliant compilers have to: it is a C++ language feature.
GCC , to name one, supports it.
"In testa che avete, Signor di Ceprano?"
-- Rigoletto
|
|
|
|
|
I mean in same way as Turbo C++ did. For C# and all are basically java. They don't do multiple inheritance as effortlessly as Turbo C++ use to.
|
|
|
|
|
Hi!
I cannot compile this program.
Please help me because I have no idea how I do it, my teacher spent this activity and I found it very difficult.
After completing an option, the program must return to the initial menu until "0" is pressed.
<pre>
#include <stdio.h>
#include <conio.h>
int main()
{
system("CLS");
int item [8][5]; int op,vaga,i,j;
float value[8][5]; float box = 0;
printf("\n");
printf ("|*--------------------Welcome to the restaurant! -------------------- *|\n");
printf("\n");
printf(" [1] Open table\n");
printf(" [2] Launch expense\n");
printf(" [3] Close table account\n");
printf(" [4] Check Cashier\n");
printf(" [0] Exit\n");
printf("\n Choose an option:");
scanf ("%d",&op);
switch (op)
{
case 0:
printf("Leaving the Program");
break;
case 1:
printf ("Enter table number:");
scanf("%d");
if (item>=-1) printf("%d table open");
else if (item<= 0)
printf("Error that table is not free. Choose another table!");
function_open_table();
break;
case 2:
printf("Value launched on table");
printf("Placement amount: %d");
scanf("%d");
printf("Total value of the table:R$ ");
scanf("%f");
if (table>=-0) printf("%d table open");
else if (table<= -1)
printf("Error that table is not open!");
function_Launch_dispatch();
break;
case 3:
printf("The table has just been closed");
break;
case 4:
printf("The total value in the box: R$%.2f");
break;
default:
printf("option invalidates!");
}
return 0;
}
|
|
|
|
|
What error(s) do you get from the compiler?
|
|
|
|
|
At a quick glance, you have this
scanf ("%d",&op);
and you have this
scanf("%d"); the second one is missing the variable where you want the data to go.
Similarly with the printf() statements
printf("Placement amount: %d"); you're specifying an integer to be output with the %d, there isn't a corresponding value.
printf("Placement amount: %d", value);
"the debugger doesn't tell me anything because this code compiles just fine" - random QA comment
"Facebook is where you tell lies to your friends. Twitter is where you tell the truth to strangers." - chriselst
"I don't drink any more... then again, I don't drink any less." - Mike Mullikins uncle
|
|
|
|
|
I managed to compile only that now what I'm having difficulty is to make the program return to the menu at the end of the command so that new commands are given and the values are saved in the box.
(I'm sorry if my English is bad, I am translating to clear my doubts here.)
#include <stdio.h>
#include <conio.h>
int main()
{
int item [8][5]; int i,j,op;
float value[8][5]; float box = 0; int table();
printf("\n");
printf ("|*--------------------"|- Welcome to the restaurant! -------------------- *|\n");
printf("\n");
//Menu
printf(" [1] Open table\n");
printf(" [2] Launch expense\n");
printf(" [3] Close table account\n");
printf(" [4] Check Cashier\n");
printf(" [0] Exit\n");
printf("\n Choose an option:");
scanf ("%d",&op);
switch (op)
// should receive only one character which are the (1,2,3,4,0) without the need to press the enter any other number should be disregarded.
{
case 0:
printf("Leaving");
break;
case 1:
printf ("Enter table number: ");
scanf ("%d",&i,&j);
if (item <=-1) //item=-1 closed table
printf("open table %d \n ",i,j);
else //////
if (item>=0) //item =0 closed table
printf("This table is not free. Choose another table!");
break;
case 2:
printf ("Enter table number:");
scanf ("%d",&i,&j);
if (item <=0) //item =0 open table
printf ("Enter with the value to be launched on the table %d: R$ \n ",i,j);
scanf ("%.2f",&i,&j,&value);
printf(" Value launched on the table.\n ");
printf("Amount posted: \n"); // show how many postings were on the table
printf(" Total table value: %.2f \n",value);
(item >=-1);
printf(" Error: This table is not open! \n"); //item=-1 table closed
break;
case 3:
printf ("Enter table number:");
scanf ("%d",&i,&j);
printf(" The table has just been closed \n");
printf(" Amount of postings: \n"); // show how many postings you had on the table
printf(" Total value of the table %d e' R$ %.2f \n",i,j,value);
break;
case 4:
printf("The total value in the box: R$"); // should show the total value you have in the box of all tables that have already been opened
scanf("%f",&box);
break;
}
return 0;
}
|
|
|
|
|
To return to the menu I'd imagine that you would need some type of loop construct, like a while() loop, or a do-while loop. Your formatted inputs and outputs don't look right;
Member 14984894 wrote: scanf ("%d",&i,&j); you're asking for a single integer (the '%d'), but you have 2 variables there i and j. What are you expecting to happen? Same goes for some of the printf() statements.
"the debugger doesn't tell me anything because this code compiles just fine" - random QA comment
"Facebook is where you tell lies to your friends. Twitter is where you tell the truth to strangers." - chriselst
"I don't drink any more... then again, I don't drink any less." - Mike Mullikins uncle
|
|
|
|
|
I left & i, & j because it is the line and the row where the tables will be chosen. I thought that this way I would be storing the tables that were chosen
|
|
|
|