I am relatively new at programming, and I'm having trouble figuring out how to loop through an array until the counter finds zero, and when it finds zero once, performs an action and exits the loop. Here is the loop I have so far:
for (int i = 0; i<13; i++)
{
if(pHand[i] == 0)
{
pHand[i] = deal(numArray);
printf("%d\n", i);
printHand(pHand, "Your");
}
}
Currently, this loops through the array until it finds zero, calls deal(), prints the value of pHand, and then loops back through the same sequence until i=0. Please help. I am completely stumped on how to fix this.
The break statement can be used to exit an enclosing loop (e.g., a while, do-while, or for) or switch.
for (int i = 0; i<13; i++)
{
if(pHand[i] == 0)
{
pHand[i] = deal(numArray);
printf("%d\n", i);
printHand(pHand, "Your");
break;
}
}
// code will continue executing here if the for loop condition becomes
// false (i is 13) or if the break statement is reached.
In your code, if you encountered ZERO value cell, you just call "deal" function and printf, but you don't exit the loop, your are continuing to the next iteration.
In order to exit the loop, add "break" statement in the "if" scope and you will go out the loop once you fulfill the condition.
Some consider break to be harmful. I've used it plenty, but some people have issues with it. If you wanted to avoid using break, you could do the following:
int i = 0;
char finished = 0;
while (i < 13 && !finished)
{
if(pHand[i] == 0)
{
pHand[i] = deal(numArray);
printf("%d\n", i);
printHand(pHand, "Your");
finished = 1;
}
i++;
}
You could also rework it to use do-while. Some would say that this kind of solution is a little nicer, semantically.
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
}
Basically, I am having issues with the while loop below in part of my program.
Here is the section of code I'm having issue with:
char *nameOfTheCommand;
char *arrayArgs[500];
//track for redirection. If set, gives position of the file name. Else it equals zero
int redirectionCheck=0;
arrayArgs[0]=token;
int i;
i=0;
//While still arguments to take in, do this
while(arrayArgs[i]!=NULL)
{
i++;
arrayArgs[i]=strtok(NULL, " \n");
if(strcmp(arrayArgs[i], "<")==0)
{
redirectionCheck=i;
}
}
All I want the code to do is loop through strtok and set it equal to arrayArgs[i]. If the strtok happens to pull out a "<" symbol, then I want redirection to be set to i.
This seems simple enough. Yet, if I include the if statement in the while loop, it seem that the while loop exits and no code after that executes. I put a printf statement after the while loop and it won't print anything, that is how I know that it is like the while loop just stops everything else running after it immidiately.
Yet, if I don't include the if statment, my code runs fine.
Can someone please explain to me why this if statment seems to be causing my while loop to not act correctly? As in, it seems the while loop just wont execute stuff after it if I include it. Thanks for any information.
In the final iteration of the loop, you pass NULL to strcmp. This can be avoided by rearranging the loop:
i = 1;
// read subsequent tokens
while((arrayArgs[i] = strtok(NULL, " \n")) != NULL)
{
if(strcmp(arrayArgs[i], "<") == 0)
{
redirectionCheck = i;
}
i++;
}
but I would also add a check on the value of i.
Beginner in C and running into a problem with a function that initializes an array. Compiled in Code:Blocks 16.01 on Windows 10. Specific code I'm having issues with is:
void initAuction(float auction[2][MAXAUCTIONITEMS]) {
int i;
for (i = 0; i < MAXAUCTIONITEMS; i++) {
auction[1][i] = -1;
printf("\n%f\t%d\n", auction[1][i], i);
};
for (i = 0; i < MAXAUCTIONITEMS; i++) {
auction[2][i] = 0;
printf("\n\n%f\t%d", auction[2][i], i);
}
printf("\n%f\n", auction[2][70]);
return;
}
I've set up print statements to see how far I'm getting before the crash and I make it to the second for loop but it crashes at i=140. If I change the constant (which is equal to 1000) then the highest I can set it to without crashing is i<84 oddly enough. What would cause the termination status -1073741819 mid loop when the first row initialized no problem but row 2 chooses to crash at around i=140.
I've tried searching on google and here and it seems the termination code isn't a very specific code since I've seen solutions from needing a return statement, trying to access something that doesn't exist, etc. Really lost.
The valid indices are auction[0][*] and auction[1][*].
You are setting elements of the array beyond its boundaries: the initial dimension of auction is 2, the only valid values for this index are 0 and 1.
You can fix and simplify the code this way:
void initAuction(float auction[2][]) {
for (int i = 0; i < MAXAUCTIONITEMS; i++) {
auction[0][i] = -1;
auction[1][i] = 0;
}
}
Note that the second dimension is not part of the type of auction, it is ignored by the compiler.
I have an array (nchar[12]) and I wrote this code to print it as vertical columns composed of "X"'s.
I first wrote a version with an accumulator and a while-loop and it worked fine, but it only could print colums as long as a given limit.
Then I tried to write it as a state machine, but the output is just an endless series of blank spaces.
I declared status as an int and assigned a value of 1 to it, then:
while (status = 1) {
for (i = 1; i <= 12; ++i) {
status = 0;
if (nchar[i] > 0) {
printf(" X");
--nchar[i];
status = 1;
}
else
printf(" ");
}
It should stop when it doesn't find any value to print for the last processed line, but it just goes on forever and I don't understand why.
The loop never ends because = is the assignment operator not == which is the comparision operator. You probably want
while (status == 1)
Or simply
while (status)
instead of
while (status = 1)
Also if you have an array declared as
type nchar[12];
then the valid indices for it start from 0 and end at 11. So, your loop should start with i=0 and should loop until i<12 becomes false.
I am confused by the for(;;) construct. I think it is a form of shorthand for an unlimited for loop but I can't be sure.
Here is the code:
for(;;)
{
//whatever statements
}
Your guess is correct; it's an infinite loop.* This is a common C idiom, although many people (including me) believe the following to be less cryptic:
while (1) { whatever statements; }
* It's infinite assuming there are no break/return/etc. statements inside the loop body.
It's an un-terminated loop. It is sometimes written with a while:
while (1)
or even better:
while (true)
I would expect to see a break or return inside any such loop, no matter whether it is written with for or while. There has to be some abnormal control flow or it really will be an infinite loop.
Yes, that's the for C syntax with blank fields for initialization expression, loop condition and increment expression.
The for statement can also use more than one value, like this sample :
for (i=0, j=100, k=1000; j < 500 || i<50 || k==5000; i++, j+=2, k*=6) {};
Maybe one step beyond in for understanding ? =)
Yes, the expressions in the for loop are just optional. if you omit them, you will get an infinite loop. The way to get out is break or exit or so.
This statement is basically equal to:
while(1) {}
There is no start, no condition and no step statement.
As I understand it, for(;;) creates a deliberate non-exiting loop. Your code is expected to exit the loop based on one or more conditions. It was once provided to me as a purer way to have a do while false loop, which was not considered good syntax. Based on the exit condition, it is easier to dispatch to a function to handle the result, failure, warning, or success, for example.
My explanation may not be the reason someone used that construct, but I'll explain in greater detail what it means to me. This construct may be someone's "Pure C" way of having a loop in which you can serially perform multiple steps, whose completion mean something like your application has performed all steps of initialization.
#define GEN_FAILURE -99
#define SUCCESS 0
/* perform_init_step1() and perform_init_step2() are dummy
place-holder functions that provide a complete example.
You could at least have one of them return non-zero
for testing. */
int perform_init_step1();
int perform_init_step2();
int perform_init_step1()
{
return 0;
}
int perform_init_step2()
{
return 0;
}
int ret_code = GEN_FAILURE;
for(;;)
{
if(SUCCESS != perform_init_step1())
{
ret_code = -1;
break;
}
if(SUCCESS != perform_init_step2())
{
ret_code = -2;
break;
}
break;
}
If part of the initialization fails, the loop bails out with a specific error code.
I arrived at using C having done a lot of firmware work, writing in assembly language. Good assembly language programmers taught me to have a single entry point and single exit. I took their advice to heart, because their creed helped them and me immensely when debugging.
Personally, I never liked the for(;;) construct, because you can have an infinite loop if you forget to break; out at the end.
Someone I worked with came up with do..until(FALSE), but the amount of proper C furvor this caused was not to be believed.
#define GEN_FAILURE -99
#define SUCCESS 0
/* perform_init_step1() and perform_init_step2() are dummy
place-holder functions that provide a complete example.
You could at least have one of them return non-zero
for testing. */
int perform_init_step1();
int perform_init_step2();
int perform_init_step1()
{
return 0;
}
int perform_init_step2()
{
return 0;
}
int ret_code = GEN_FAILURE;
do
{
if(SUCCESS != perform_init_step1())
{
ret_code = -1;
break;
}
if(SUCCESS != perform_init_step2())
{
ret_code = -2;
break;
}
}
until (FALSE);
This runs once, no matter what.