I'm doing a program where you can add items for a shop. Before the user can be allowed to do that, theres a subfunction which will ask for a user and password; if they are correct then allows the user to add any item as far he wants, also there's a variable called count if this variable goes to "0" should print and error and several noises and after that bring back the user to the main function and close the program.
The problem is after the condition is 0 the noises and error shows correctly but the program backs to ask for user and password and the count variable starts at 0 and every try rest the int...-1 -2 -3.
So I try this, even its working I think it's a pretty bad way:
The call
while(credencial !=1)
{
credencial=checkout(&counter);
if(counter==88)
{
goto error;
}
}
Function
int checkout(int *counter){
char login[10], password[10];
int c = 0;
printf("\nNumero de intentos restantes: %d\n", *counter);
puts("\nIntroduce un usuario con credenciales: \n");
scanf("%10s", login);
puts("\nIntroduce la contraseña: \n");
scanf("%10s", password);
(*counter)--;
if (*counter == 0)
{
system("cls");
for (c = 0 ;c < 3 ;c++)
{
puts("\n\aACCESSO DENEGADO\n");
}
*counter=88;
}
if (strcmp(strlwr(login), "admin") ==0 && strcmp(strlwr(password),"admin")==0)
{
return 1;
}
}
You can either use break; to leave the while loop (then you need to have a way to decide after the loop if it is a normal or error exit, for example by checking credential value) or add an additional condition to the while, like
while(credentiel != -1 && !abort) {
....
if (*counter == 88)
abort = true;
Or cou can use the goto, it is not that bad. Especially if the error label is used by other places as well.
BTW: I don't know why you mix out variables with return types and why you use 88 as a special number. Thats all quite untypical. You can keep the counter in the outer loop and make the crendtial check stateless, let it return pass/fail only. It is more common to use negative values (-1) for special "signals". In all cases make sure nobody can force the special value with other tricks.
Related
Can you break out of an if statement or is it going to cause crashes? I'm starting to acquaint myself with C, but this seems controversial. The first image is from a book on C
("Head First C") and the snippet shows code written by Harvard's CS classes staff. What is actually going on and has it something to do with C standards?
breaks don't break if statements.
On January 15, 1990, AT&T's long-distance telephone system crashed, and 60,000 people lost their phone service. The cause? A developer working on the C code used in the exchanges tried to use a break to break out of an if statement. But breaks don't break out of ifs. Instead, the program skipped an entire section of code and introduced a bug that interrupted 70 million phone calls over nine hours.
for (size = 0; size < HAY_MAX; size++)
{
// wait for hay until EOF
printf("\nhaystack[%d] = ", size);
int straw = GetInt();
if (straw == INT_MAX)
break;
// add hay to stack
haystack[size] = straw;
}
printf("\n");
break interacts solely with the closest enclosing loop or switch, whether it be a for, while or do .. while type. It is frequently referred to as a goto in disguise, as all loops in C can in fact be transformed into a set of conditional gotos:
for (A; B; C) D;
// translates to
A;
goto test;
loop: D;
iter: C;
test: if (B) goto loop;
end:
while (B) D; // Simply doesn't have A or C
do { D; } while (B); // Omits initial goto test
continue; // goto iter;
break; // goto end;
The difference is, continue and break interact with virtual labels automatically placed by the compiler. This is similar to what return does as you know it will always jump ahead in the program flow. Switches are slightly more complicated, generating arrays of labels and computed gotos, but the way break works with them is similar.
The programming error the notice refers to is misunderstanding break as interacting with an enclosing block rather than an enclosing loop. Consider:
for (A; B; C) {
D;
if (E) {
F;
if (G) break; // Incorrectly assumed to break if(E), breaks for()
H;
}
I;
}
J;
Someone thought, given such a piece of code, that G would cause a jump to I, but it jumps to J. The intended function would use if (!G) H; instead.
This is actually the conventional use of the break statement. If the break statement wasn't nested in an if block the for loop could only ever execute one time.
MSDN lists this as their example for the break statement.
As already mentioned that, break-statement works only with switches and loops. Here is another way to achieve what is being asked. I am reproducing
https://stackoverflow.com/a/257421/1188057 as nobody else mentioned it. It's just a trick involving the do-while loop.
do {
// do something
if (error) {
break;
}
// do something else
if (error) {
break;
}
// etc..
} while (0);
Though I would prefer the use of goto-statement.
I think the question is a little bit fuzzy - for example, it can be interpreted as a question about best practices in programming loops with if inside. So, I'll try to answer this question with this particular interpretation.
If you have if inside a loop, then in most cases you'd like to know how the loop has ended - was it "broken" by the if or was it ended "naturally"? So, your sample code can be modified in this way:
bool intMaxFound = false;
for (size = 0; size < HAY_MAX; size++)
{
// wait for hay until EOF
printf("\nhaystack[%d] = ", size);
int straw = GetInt();
if (straw == INT_MAX)
{intMaxFound = true; break;}
// add hay to stack
haystack[size] = straw;
}
if (intMaxFound)
{
// ... broken
}
else
{
// ... ended naturally
}
The problem with this code is that the if statement is buried inside the loop body, and it takes some effort to locate it and understand what it does. A more clear (even without the break statement) variant will be:
bool intMaxFound = false;
for (size = 0; size < HAY_MAX && !intMaxFound; size++)
{
// wait for hay until EOF
printf("\nhaystack[%d] = ", size);
int straw = GetInt();
if (straw == INT_MAX)
{intMaxFound = true; continue;}
// add hay to stack
haystack[size] = straw;
}
if (intMaxFound)
{
// ... broken
}
else
{
// ... ended naturally
}
In this case you can clearly see (just looking at the loop "header") that this loop can end prematurely. If the loop body is a multi-page text, written by somebody else, then you'd thank its author for saving your time.
UPDATE:
Thanks to SO - it has just suggested the already answered question about crash of the AT&T phone network in 1990. It's about a risky decision of C creators to use a single reserved word break to exit from both loops and switch.
Anyway this interpretation doesn't follow from the sample code in the original question, so I'm leaving my answer as it is.
You could possibly put the if into a foreach a for, a while or a switch like this
Then break and continue statements will be available
foreach ([1] as $i) if ($condition) { // Breakable if
//some code
$a = "b";
// Le break
break;
// code below will not be executed
}
for ($i=0; $i < 1 ; $i++) if ($condition) {
//some code
$a = "b";
// Le break
break;
// code below will not be executed
}
switch(0){ case 0: if($condition){
//some code
$a = "b";
// Le break
break;
// code below will not be executed
}}
while(!$a&&$a=1) if ($condition) {
//some code
$a = "b";
// Le break
break;
// code below will not be executed
}
I'm working on this project where a user has to guess a word (wordToGuess) and he has a number of attempts.
The problem is that the variable "wordToGuess" loses its value when the code arrives in the point marked ("HERE LOSES ITS VALUE). I don't know how to solve this problem, I've tried in many ways. Thank u for your help!
(checkExistence is a function that checks if the word is present in the dictionary)
void newGame(node* head){
char wordToGuess[10];
char attempt[10];
int numberOfAttempts = 0;
if (scanf("%s" , wordToGuess) != 1){
printf("error1");
}
getchar();
if (scanf("%d", &numberOfAttempts) != 1){
printf("error2");
}
getchar();
while(numberOfAttempts > 0){
if (scanf("%s", attempt) != EOF){
if (attempt[0] != '+'){
if (checkExistence(head, attempt) == false){
printf("not_exists\n");
}else{
if (strcmp(wordToGuess, attempt) == 0){
printf("ok\n");
return;
}else{
//code
numberOfAttempts--;
}
}
}else{
if (attempt[0] == '+' && attempt[1] == 's'){
//HERE LOSES ITS VALUE
}else if (attempt[0] == '+' && attempt[1] == 'i'){
//other code
}
}
}else{
printf("ko");
return;
}
}
return;
}
Here a test case:
2rj9R (wordToGuess)
18 (numerAttemps)
DP3wc (attempt)
7PGPU (attempt)
2rz9R (attempt)
+print_list (from this point I lose the value of wordToGuess)
2rj9R (attempt)
As the others have point, you're probably causing a buffer overflow in your attempt buffer which overwrites your wordToGuess buffer since your attempt and wordToGuess buffer is stored like this in your memory:
<attempt buffer> | <word To Guess>
You have two possible fixes for this (as the comments have said...):
A little fix would be to set a limit of how many characters should be read from stdin to scanf like this:
scanf("%9s" , wordToGuess) // you need 9, because your buffer can store up to
// 10 bytes but don't forget that `scanf` is also
// addinng `\0` for you!
and don't forget to flush the rest of the user input if you want that the user should be only able to insert at most 9 characters!
Increase the buffer size of your attempt (and wordToGuess) buffer but also add those read-limits for scanf which is described in the first point.
At the indicated point of the code where wordToGuess appears to lose its value, it is a dead variable. If you're looking at optimized code in a debugger, you may find that the variable doesn't exist there any more.
At a given point in a program, a dead variable is one which is never used past that point. All control flows out of that point reach the termination of that code, without ever using the variable again. Simple example:
{
int x = 3;
// x is live here: there is a next reference
printf("%d\n", x);
// x is now dead: it is not referenced after the above use
printf("foo\n");
}
In the generated code, the compiler may arrange to re-use the resources tied to a dead variable as soon as it is dead: give its register and memory location to something else.
In the debugger, if we put a breakpoint on the printf("foo\n") and try to examine x, we might get a strange result.
To have the best chance of seeing the expected result (that x still exists, and is retaining its most recent value), we have to compile with optimizations disabled.
I am having issues making this nested while loop to work in this terminal application.I am trying to build a system with Clients and an Admin with certain
capabilities for each.
scanf("%s",sign_in_choice);
int admin_result, client_result, number_of_clients, x, c;
admin_result = strcmp(sign_in_choice,"Admin");
client_result = strcmp(sign_in_choice,"Client");
char client_choice[10];
int client_username;
int client_password;
int client_choice_log_in, client_choice_register;
while(x == 0)
{
if(admin_result == 0 )
{
printf("Welcome to Admin Page\n");
}
else if(client_result == 0)
{
printf("Testing");
while(c == 0)
{
//START OF CLIENT PAGE
printf("Welcome to Client Page\n");
printf("Register or Log in?\n");
scanf("%s",client_choice);
client_choice_log_in = strcmp(client_choice,"Login");
client_choice_register = strcmp(client_choice,"Register");
if(client_choice_register == 0)
{
// REGISTER
reg(client_number);
// END OF REGISTER
}
else if(client_choice_log_in == 0)
{
//LOG IN
printf("Please enter your credentials\n");
printf("Enter your afm:\n");
scanf("%d", &client_username);
printf("Enter your year of birth:\n");
scanf("%d", &client_password);
login(client_username, client_password, client_number, client_verify);
//END OF LOG IN
// PERSONAL CLIENT PAGE
i = client_number;
printf("Welcome to your Personal Client Page: %s \n", client_array[i].client_name);
printf("Verify : %d\n",client_verify);
// END OF PERSONAL CLIENT PAGE
}
else
{
printf("Invalid input.Please try again.\n");
scanf("%s",client_choice);
client_choice_log_in = strcmp(sign_in_choice,"Login");
client_choice_register = strcmp(sign_in_choice,"Register");
}
// END OF CLIENT PAGE
}
}
else
{
printf("Invalid input.Please try again.\n");
scanf("%s",sign_in_choice);
admin_result = strcmp(sign_in_choice,"Admin");
client_result = strcmp(sign_in_choice,"Client");
}
}
If I remove the while( c == 0) loop the code works.It allows me to select either Login or Register and continue.If I leave the loop as is when I input 'Client' the program goes into an infinite loop printing 'testing'.In addition if I enter 'Admin' the program again goes into infinite loop printing 'welcome to admin page' although the while loop appears later in the program.At least in my understanding the while(c == 0) loop should only affect what is contained within it not outside it,at least in this particular case.I am using Devc 5.11.Also I tried compiling and running through vs code and codeblocks and in those cases the program doesn't even go beyond the 'Admin Client' choice in the beginning.As you can see I have some functions and and structs not shown here but they should not affect my problem since everything works fine until I try to make this nested while loop work.
c is uninitialized, so it has an undefined behavior, which depends on your compiler. You will need to initialize your variables. And to make sure you do not have infinite loops, you must handle the case when you have a client and c is not 0. You can set it to 0, or something, just don't leave there an else which will always be entered, will not change c and will have a loop inside it, which depends on the value of c.
Maybe I missed it, but I do not see how you set the value of c inside the while (c == 0) loop, so if c happens to be 0, that inner loop seems to be infinite as well.
here is a part of my code. When I run my code, it's requesting an input from user and then matching it with another integer which recorded in my structure. When user input is matching, it is working correct. But when user enters a wrong input, it gives a segmentation fault. In where, I should make changes on my code?
long int userInput,endCheck; // Input from user
int flag=0; // check for match
int status;
int c; // controlling ctrl+D
int position= 999; // position of where the user input and data matched
LABEL:
printf("\n\t---------------------------------------\n");
printf("\n\nPlease enter the student ID which you want to find(3-times CTRL+D for finish):\n");
scanf("%d",&userInput);
if( (c=getchar()) == EOF){
exit(0);
}
for(i=0;i<lines,flag==0;i++){
if(index[i].id == userInput){
position=i;
flag=1;
}else{
position=999;
}
}
if(flag==0){
printf("id not found");
}
studentInfo info; // for storing the information which we will take between determined offsets
if(position!= 999){
if ( (pos = lseek(mainFile,index[position].offset , SEEK_SET)) == -1)/*going to determined offset and setting it as starting offset*/
{ perror("classlist"); return 4; }
while ( (ret= read(mainFile,&info, sizeof(info))) > 0 ){
printf("\n\nStudent ID: %d, Student Name: %s\n\n",info.id,info.name);
break;// to not take another students' informations.
}
}
flag=0;
goto LABEL;
printf("Program is terminated");
The right way to do that loop with the unwanted comma is like this. When you find the right index[i].id you can exit the loop early by using break.
for(i=0;i<lines;i++){
if(index[i].id == userInput){
position=i;
flag=1;
break;
}
}
You don't need the else branch as position is set to 999 from the outset of the code. But really you shouldn't use position in this fashion. What if you have more than 999 records? You're already using flag to identify if you've set position to a valid value. You should replace any instance of if(position!= 999) with if(flag).
Or since position is a signed int, you could use a negative value and ditch the flag.
The reason can be the fact that you are reaching an index that doesn't exist in the end of cycle, in the moment of the "if" statement with iterator "i".
Or in the last if, where you access a "position" index of the array. Check those limits.
Also, try GDB, is useful for solving this kind of problems.
I have a work to do in which I have to keep a loop inside the function expecting the following parameters:
-"i" to insert
-"s" to search
-"q" to quit
How do I keep this loop? I've looked up some options and it seems to be possible using a while or a switch, but I am not sure which is the best way to read those chars (with a fscanf perhaps?). I am also not sure how to read the things after the parameter "i" as the input would be "i word 9", so after detecting the i to insert I have to read a string and an int.
Anyone has any idea how to do this? I am sorry is this seems simple, but I am new to programming.
edit: Here is what I have so far
while (loop) {
fscanf(stdin,"%c",&par);
if (strcmp(&par,"i")){
scanf("%s %d",palavra,p);
raiz = insere(raiz,&palavra,p);
}
else if (strcmp(&par,"b")){
scanf("%s",palavra);
busca(raiz,&palavra);
}
else if (strcmp(&par,"q"))
loop = 0;
}
edit 2: This is what I have now, I am having problems reading the string and integer when the parameter is i, somehow it crashes the function
while (1) {
c = getchar();
if (c == 'f')
break;
else if (c == 'i'){
fscanf(stdin,"%s",&palavra);
scanf("%d",&p);
raiz = insere(raiz,palavra,p);
}
else if (c == 'b') {
scanf("%s",palavra);
busca(raiz,palavra);
}
}
Thanks in advance!
The code you have doesn't look too bad compared to what I believe you want. You can replace the "while (loop)" with "while (1)" and then your exist code "loop = 0;" with "break;" which is a bit more standard way of doing things. Also "fscanf(stdin..." is the same as "scanf(..." ... scanf will read from stdin by default. You might want to check the docs for strcmp because it returns 0 for an exact match and I don't think that will do what you want in your 'if' statements. You should be able to use scanf to read in the values you want, is it giving you an error?
You are using 3 separated scans. That means you can't input this "i word 9", but input one command or parameter at the time separated by EOL(pressing enter).. i, enter, word, enter, 9, enter ... Then the function should actually get further in those "if"s. With those scans you also should consider printing information about expected inputs ("Choose action q/i/f")
And I would recommend using something to test those inputs.
if (scanf("%d", &p) == 0) {
printf("Wrong input");
break;
}