my code is to show a few questions that are contained in a file when the user select the option to.
But i'm facing this problem: After de program show 2 questions it always shows only the last line of my file.
I don't know what is wrong.
Here is the code (only de case 1 of the switch menu)
printf("\nThe selected subject was Geology");
while(continue != 0)
{
r=0;
srand(time(NULL));
r = rand()%7;
printf("\n%d", r);
if(r==0)
r=1;
for(i=0; i<r; i++)
fscanf(arqgeo, "%s %s %s %s %s %c", question, alta, altb, altc, altd, &respa);
printf("\n\n%s \n%s \n%s \n%s \n%s \n", question, alta, altb, altc, altd);
printf("Enter alternative: ");
setbuf(stdin,NULL);
scanf("%c",&resp);
if(arqgeo == NULL)
{
printf("An error has occurred\n");
printf("Contact the developers !!!\n");
} else {
if(resp == respa)
{
printf("You're right!!!\n");
pont++;
printf("\n\n\n");
} else {
puts("\a");
printf("You missed!!!");
printf(" The correct answer is: %c", respa);
erro++;
printf("\n\n\n");
}
}
printf("Do you wish to continue? Enter a number other than 0 ");
scanf("%d", &continue);
}
break;
where: "respa" is the right answer, "alta-d" is the alternatives, "resp" is the answer of the user, "arqgeo" is the file that contain the questions
--sorry for the bad english--
Adding to Jonathan Leffler's and my earlier comments.
Each iteration reads some lines from the file. The first iteration works fine but the second picks up where the first left of so you blow through a lot of lines quickly.
What is needed is to rewind the file to the beginning at the start of each iteration. The fseek() function will do this for you. The trouble is you will also repeat questions if rand() % 7 returns the same number again. A static array initialized to all zeros to check if an question has already been used would be needed. Or use a bit map. rewind() is a special case of fseek() that returns to the start of the file (think old-style magnetic tapes).
Also, if(arqgeo == NULL) is wrong as the value of apqgeo does not change on troubles. Look at the ferror() and feof() functions to test eof/error conditions. scanf() also returns the count of the fields converted with EOF indicating some type of trouble where these "f" functions would be useful. At any rate returns of <= 0 indicate time to get out (EOF is usually (-1)).
While it's annoying to have to test each and every input statement for troubles it is good practice. If you don't you need to be hyper alert for behaviors caused by silent errors.
Challenge: arrange question file and and it's handling to get rid of the r == 0 test. If you have 8 questions you really don't need it.
Related
So first of all, here is the relevant part of my code and I will then proceed to explain.
scanf("%d", &option);
fflush (stdin);
//Insert
if(option==1){
printf("enter name\n");
fgets(name,MAX_SIZE,stdin);
printf("name: %s",name);
char str[]="alpha one";
fd=open(argv[1],O_RDWR|O_CREAT|O_TRUNC,S_IRWXU);
write(fd,str,strlen(str) + 1);
pcounter = updateCounter(pcounter, str);
lseek(fd, 0, SEEK_END);
char passpos[5];
sprintf(passpos,"%d",pcounter); //int to string
fd=open(argv[2],O_RDWR|O_CREAT|O_TRUNC,S_IRWXU);
write(fd,passpos,3);
}
The reason I am redefining str as "alpha one" is because, even though I am trying to clear the buffer after scanf with fflush, fgets still gets a "/n" as input (I assume) and when I print "name" I get blank. Any help with this would be valuable even though its not the main issue.
This is all in a loop, so I thought if I keep pressing "1" when prompted, my binary file should look like: alpha onealpha one alpha one ...
However, it looks like: alpha one, no matter how many times I press 1.
Same thing with my other binary file that stores the counter, it gets updated correctly so after the first loop it has "9" stored, but after the second only "18" when I would be expecting: 9 18
I tried removing lseek altogether, and also setting it to CURR instead of END. Same results.
I have max warnings turned on and I get none when compiling.
I should point out that this is homework so I cannot use fopen etc, only system commands.
Here is the updateCounter function in case someone wants to run it:
int updateCounter(int pcounter, char *str){
int m,charcount = 0;
for(m=0; str[m]; m++) {
charcount ++;
}
printf("chars: %d \n", charcount);
pcounter = pcounter + charcount;
printf("pcounter = %d \n", pcounter);
return pcounter;
}
Thanks for any help.
You are opening the file with the O_TRUNC flag. This tells the computer to delete all the data from the file. If you don't want to delete all the data from the file then don't use O_TRUNC.
Also, make sure to seek to the end of the file before writing any data. Otherwise, you will overwrite the data at the beginning of the file instead. You could use lseek to seek to the end, or you could also use the O_APPEND flag when opening the file to automatically seek to the end.
I've been introduced to C this year as part of my degree, during which I have to write simple programs and test them to be idiot-proof by running them over and over again, putting nonsense variables in, etc. and I had an idea to write a program with the ability to restart itself without having to run the program again.
I've tried writing a program to perform this function (which turned out to be harder than I first thought) and I now have it working, albeit using a goto function that are frowned upon. Now the only problem I have is a while loop to check for nonsense input, that seems determined to run at least once ignoring a prompt for a valid input.
Please could someone give me an idea why this is happening? (My compiler is Dev-C++ 4.9.9.2)
int main (void)
{
mainprogram:
printf("\nPROGRAM START\n");
//code copied from an exam, to check that the program performs a function
//when ran through again
int i,j,k;
printf("Please enter 7:");
scanf("%d",&i);
printf("Please enter 4:");
scanf("%d",&j);
printf("Please enter 0:");
scanf("%d",&k);
//this is to check that the program will take input when it is restarted
do {
switch (i%j) {
case 3:
i--;
k*=i;
break;
case 2:
i--;
k+=i;
default:
i--;
k++;
break;
}
printf("i is %d k is %d\n",i,k);
} while (i>0);
//end of copied code
char prompt='y';
printf("\nRestart program?");
scanf("%c",&prompt);
while (prompt != 'y' && prompt != 'Y' && prompt != 'n' && prompt != 'N')
{
//this is the problem section, when entering nonsense input, the error messages
//display twice before pausing for input, and when restarted, the program does
//run again but displays the error messages once before pausing for input
printf("\nERROR: INVALID INPUT");
printf("\n\nRestart program?");
prompt='y';
scanf("%c",&prompt);
}
if (prompt == 'y' || prompt == 'Y')
{
goto mainprogram;
}
//
return 0;
}
while(1){ //parent
printf("\n\nRun program?");
scanf("%c",&prompt);
if (prompt == 'n' || prompt == `N`)
{
printf("\nEXITINT")
return 0;
}
int i,j,k;
printf("Please enter 7:");
scanf("%d",&i);
printf("Please enter 4:");
scanf("%d",&j);
printf("Please enter 0:");
scanf("%d",&k);
switch (i%j)
{
case 3:
i--;
k*=i;
break;
case 2:
i--;
k+=i;
break;
default:
i--;
k++;
break;
}
printf("i is %d k is %d\n",i,k);
} //end while parent
//end of copied code
There are a couple of ways one could restart a program, or more generally loop over some code. Of course, all of that can be done with gotos, but that makes the code hard to understand, so C has equivalent structures for dealing with the most common patterns:
execute code if and as long as condition holds
while (condition)
{
/* code */
}
This means before executing the code, the condition is checked. If the condition holds (its value is non-zero), the code is executed and then looped back to the top. This is equivalent to:
top_of_while:
if (!condition)
goto done;
/* code */
goto top_of_while:
done:
execute code and redo while condition holds
do
{
/* code */
} while (condition)
This means execute code first and then check for a condition. If the condition holds, executed the code again. This is equivalent to:
top_of_do_while:
/* code */
if (condition)
goto top_of_do_while;
iteration
for (initialization; condition; iteration)
{
/* code */
}
This is a kind of while loop that happens a lot, in which there is an initialization, followed by a while loop, which on the bottom changes a variable to form some sort of iteration. This is equivalent to:
initialization;
while (condition)
{
/* code */
iteration;
}
To restart a program, most likely you want the do-while loop, since for sure you know that the program has to execute once. However, by properly initialization the condition variable of a while loop, you can also ensure that the loop is always entered the first time. It's a matter of style and your liking.
Where to actually use goto
Many people would tell you to never use goto. This roots from the fact that overuse of goto had led to a great number of overly complicated programs, the so-called spaghetti code. The reason is that it's hard to build a mental model of a program where the execution can jump around to any other part.
However, gotos are actually very useful in C, without which error-handling becomes a huge pain in the ... neck. This useful usage of goto is for handling errors and cleanup, which are always in the bottom of the function. An example would be:
int *read_nums(int n)
{
int *array, i;
array = malloc(n * sizeof *array);
if (array == NULL)
goto exit_no_mem;
for (i = 0; i < n; ++i)
if (scanf("%d", &array[i]) != 1)
goto exit_bad_input;
return array;
exit_bad_input:
free(array);
exit_no_mem:
return NULL;
}
This way, the code is not cluttered (much) with error handling and cleanup is done very nicely based on how far the function has executed.
I just reformatted your code and indeed # herohuyongtao is right, the break; for case 2 has moved at the end of default which is not useful there.
But there's something really shocking in your code, is that you use a goto. Just remember that rule: WHENEVER YOU USE GOTO, THERE'S A BETTER WAY TO DO IT!
#include <stdio.h>
short int read_input (void) {
printf("\nPROGRAM START\n");
//code copied from an exam, to check that the program performs a function
//when ran through again
int i,j,k;
printf("Please enter 7:");
scanf("%d",&i);
printf("Please enter 4:");
scanf("%d",&j);
printf("Please enter 0:");
scanf("%d",&k);
//this is to check that the program will take input when it is restarted
do {
switch (i%j) {
case 3:
i--;
k*=i;
break;
case 2:
i--;
k+=i;
break; // break at the right spot
default:
i--;
k++;
}
printf("i is %d k is %d\n",i,k);
} while (i>0);
// getchar(); // could be inserted here (discards one char)
// fflush(stdin); // could also do the job (discards all remaining chars in buffer)
char prompt='y';
// here the design choice is to let the user input whatever
// and not updated the output until the right character is given
// which will hide how many wrong chars has been previously in the buffer.
printf("\nRestart program? ");
do {
prompt = getchar();
} while (prompt != 'y' && prompt != 'Y' && prompt != 'n' && prompt != 'N');
if (prompt == 'y' || prompt == 'Y')
return 1;
return 0;
}
int main() {
while (read_input() == 1);
return 0;
}
Now that the code is clean, for your exact problem, you will get into a carriage return problem. Basically, what you do is get input until \n is hit, but when you hit carriage return, you actually send CR + LF. So there's one character that never gets read in the input buffer, that you need to discard.
So you should first read that SO question that sums up very well your problem, and you can either:
add a lone getchar(); just after your //end of copied code comment,
or a fflush(stdin) can do the job (cf the SO Question to learn more about it) but has been designed about flushing output buffers, not input ones,
add a fseek(stdin,0,SEEK_END); which is a bit dirty (and non-portable) but works,
or change your conditions and your use of scanf to take into account that you're actually having more chars in the buffer.
In the code I gave you, I chose the most simplistic solution, which is to discard any wrong input without printing anything.
I have a simple program, intended simply to excercise my freshly gained knowledge of C.
Please keep in mind I have attempted to solve the problem of the newline leftovers remaining in the stdin buffer using the knowledge I already have. From the problems I was having before I figured out the newline issue, I have made considerable progress in getting this program to act the way I imagine it should. This is all I have left to debug and I am stymied. While my fluency in C is marginal, at best, I believe this program can easily work without a whole lot of modification.
The program asks the user to select a tool, of 2 simple tools I have constructed. The tools themselves work fine, but the cheesy 'pick-a-letter' UI that I've built has a bug I just can't figure out. It runs through the program as expected the first time, but the second time, the variable used for tool selection gets switched to a newline character, and the user input assignment for that variable gets overwritten. Therefore the program enters the switch construct used for the tool selection variable and activates:
case '\n':
which prints an error message indicating a newline got in the way, and breaks out to return to the beginning of the menu loop. On this third time around, the program will again work as expected. It will continue to alternate like this until you exit the program via the exit provided at the end of a normal program run.
This is a direct cut and paste of pertainant code, not included are the D_Time(); and cal(); functions. Feel free to compile and run with substitue functions to see the problem firsthand if need be. Both are void functions, one is a simple 2 argument calculator and the other displays date and time via stdout. And any other suggestions you have that could improve my knowledge of C or this program would be happily accepted and are sought. I am still learning, and although pointers and calculus escape me, i will get the hang of this.
void clrnl(void)
{
while (getc(stdin) != '\n');
} /* This is the fix that kind-of but not-really fixes the '\n' problem. Without it, the error message would pop up every time i used the menu, not including the first run. */
void menu(void)
{
char tool = 1;
int r = 0;
while (r != 1)
{
printf("\n ---Select a tool.---\n\n ------------------\n\n C = Calculator\n\n D = Date/Time\n\n ");
tool = getc(stdin);
clrnl(); /* This seems to be the only line where clrnl(); has a positive effect. */
switch (tool)
{
case 'D':
printf("\n You selected Date/Time ");
D_Time();
r = 1;
break;
case 'C':
printf("\n You selected Calculator ");
cal();
r = 1;
break;
case 'd':
printf("\n You selected Date/Time ");
D_Time();
r = 1;
break;
case 'c':
printf("\n You selected Calculator ");
cal();
r = 1;
break;
case '\n':
printf("\n Error!! -- Newline character detected!! Try again.\n\n");
r = 0;
break;
default:
printf("\n Error!! -- Unknown. Try again.\n\n");
r = 0;
break;
}
}
}
int main()
{
char lx;
do
{
menu();
printf("\n Enter 'x' to exit.\n\n ");
scanf(" %c", &lx);
if (lx != 'x')
{
continue;
}
}
while (lx != 'x');
return 0;
}
Usually a context menu is not waiting for the user to press enter before getting the option while your loop does it.
In order to select the option when pressing the relevant key instead waiting for the "enter" you can call OS specific functions. Under linux reconfigure your terminal using system ("/bin/stty raw") just before entering in your while loop and than leave things as they are. Under windows consider _getch() instead of getc(stdin).
Remember that getchar() and getc() return an integer and not a char.
Doing this you have a more reactive menu and you don't need to handle the extra chars generated.
The problem is when you input your selection, for example D, and you press enter key, various things can happen. If you're on windows, your application will recieve characters: 'D', '\r', '\n'. On linux, it will be 'D','\n'. To eat up any extra input read by scanf use this simple trick:
scanf(" %c*", &lx);
The * in format string will skip your carriage return and newline characters.
This is changed code for tool selection:
printf("\n ---Select a tool.---\n\n ------------------\n\n C = Calculator\n\n D = Date/Time\n\n ");
scanf(" %c*", &tool);
This is the part for the exit input:
printf("\n Enter 'x' to exit.\n\n ");
scanf(" %c*", &lx);
You can further put more switch cases together like this to reduce code duplication:
case 'D':
case 'd':
printf("\n You selected Date/Time ");
D_Time();
r = 1;
break;
This will work, because there's no break statement after case 'D', so if this case happens, the code continues pass the case 'd': line into the corresponding code.
Build succeeds. If I open the output window, it reads:
Program ended with exit code: 0
But my program is such that this shouldn't be possible, without first having taken user input, done some stuff, and taken another user input, all in int main().
The first thing int main() does is loop through taking input for p, until the input is one (of two) desired options. So there's no way it should be able to exit immediately - it initialises p=0 and doesn't exit a while loop until p is 1 or 2.
Is there some hidden error that has allowed the build to succeed without it actually.. succeeding?
int main(){
//vars
while (TRUE){
//play computer or human?
while (!(p == 1 || p == 2)) {
printf("Single player or two player? (1/2): ");
scanf("%d", &p);
}
if (p==1) {
//play computer
}
else {
//snip
}
printf("%s won the game! Play again?", winner);
scanf("%s", playagain);
if (strncmp(playagain,"no",2)==0){
break;
}
}
return 0;
}
Print out what uppercase TRUE is defined as. I have seen some ambitious but inexperienced folks do weird things with it. You might not even be getting into your main while loop.
printf("TRUE is %d\n",TRUE);
If this is nonzero, then your problem is elsewhere.
EDIT: About to try intilizing my chars properly... [didnt work=( ]
EDIT: SOLVED (I can't answer my own question for another 7 hours...)
Thanks for the comment Brian, that's just a constant declared at the top.. (it = 20).
It turns out the error was happening because I forgot to add a next line after I took in the input name.
it's working now =D
:ENDEDIT (lol)
I've code my code below, Basically I put in the first name this is supposed to find
John
Then I put in the last name...
Locke
and as soon as I enter in "Locke" it hits this error, I feel like maybe it's the scanf and I should be using a better alternative ???
int findPatron(struct Library *lib,struct Patron **p)
{
int i;
char firstName[NAME_LENGTH], lastName[NAME_LENGTH];
printf("\nPlease enter the patron's first name: ");
scanf("%s",firstName);
printf("\nPlease enter the patron's last name: "); //this line prints...
scanf("%s",lastName); //SEGMENTATION ERROR happens here I'm pretty sure.
printf("deerrrr"); //this line never prints
for(i = 0; i<lib->totalPatrons;i++)
{
printf("checking %s %s to %s %s",lib->patrons[i].name.first,lib->patrons[i].name.last,firstName,lastName);
if((strcmp(lib->patrons[i].name.first, firstName) == 0) && (strcmp(lib->patrons[i].name.last, lastName) == 0))
{
**p = lib->patrons[i];
return 0;
break;
}
}
return 1; //No Match!
}
You're getting the segmentation fault after your scanf() statements.
If you remove everything after printf("deerrrr"); and add a \n to that output so the buffer is flushed, you'll find that it all works just fine (provided NAME_LENGTH is at least 6 given your example input).
Part of programming is knowing how to isolate and debug your problems.
Your issues are with your loop and lib struct - you're dereferencing something you shouldn't be.
SEGMENTATION ERROR happens here I'm pretty sure
this line never prints
C's printf is buffered, and only gets flushed by an explicit call to fflush or a blocking action (like scanf, which AFAIK flushes stdout as well), so the error could happen in another place. Learn to use debugger, that's the proper way of debugging C programs.