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.
Related
We need to create a binary tree which contains content of textfiles. The pointer selection_a and selection_b pointing to another textfile in the directory.
The structure of the textfiles is following:
line: Title
line: OptionA
line: OptionB
line: Text.
The first file is given as parameter while starting the program. All files should be saved at the beginning of the program. Then the text of the first file shows, and the user can input A or B to continue. Based on the selection, the text of File Option A/B is shown and the user can decide again.
The last file of a tree contains no Options: lines 2 and 3 are "-\n".
The problem is, this code only reads all the option A files of the first tree. It doesn't read in any B-Options. In the end, the program shows a memory access error.
I think the problem is that the readingRows function has no abort condition.
current->selection_a = readingRows(input_selection_a);
current->selection_b = readingRows(input_selection_b);
I know the code may be kind of chaotic, but we are beginners in programming. Hope anybody can help us to write an abort-condition.
The function should be aborted if the content of option A (line 3) is "-\n".
Here is the whole function:
struct story_file* readingRows(FILE *current_file)
{
char *buffer = fileSize(current_file);
char *delimiter = "\n";
char *lines = strtok(buffer, delimiter);
int line_counter = 0;
struct story_file *current = malloc(sizeof(struct story_file));
while(lines != NULL)
{
if(line_counter == 0)
{
current->title = lines;
}
else if(line_counter == 1)
{
char *filename_chapter_a = lines;
FILE *input_selection_a = fopen(filename_chapter_a, "r");
if(input_selection_a)
{
current->selection_a = readingRows(input_selection_a);
}
fclose(input_selection_a);
}
else if(line_counter == 2)
{
char *filename_chapter_b = lines;
FILE *input_selection_b = fopen(filename_chapter_b, "r");
if(input_selection_b)
{
current->selection_b = readingRows(input_selection_b);
}
fclose(input_selection_b);
}
else if (line_counter >= 3)
{
current->text = lines;
}
lines = strtok(NULL, delimiter);
line_counter++;
}
return current;
}
There are two items that define a terminating recursive function:
One or more base cases
Recursive calls that move toward a base case
Your code has one base case: while (lines!=NULL) {} return current;, it breaks the while loop when lines is NULL and returns current. In other words, within any particular call to your function, it only terminates when it reaches the end of a file.
Your code moves toward that base case as long as your files do not refer to each other in a loop. We know this because you always read a line, take an action according to your if-else block, and the read the next line. So you always move toward the end of each file you read.
But as you note, the issue is that you don't have a case to handle "no Options", being when lines 2 or 3 are "-\n". So right now, even though you move through files, you are always opening files in line 2. Unless a file is malformed and does not contain a line 2, your recursive call tree never ends. So you just need to add another base case that looks at whether the beginning of lines matches "-\n", and if it does, return before the recursive call. This will end that branch of your recursive tree.
Inside of your while loop, you will need code along the lines of:
if `line_counter` is `2` or `3`
if `lines` starts with your terminating sequence "-\n"
return current
else
`fopen` and make the recursive call
In the parent function that made the recursive call, it will move to the next line and continue as expected.
P.S. Make sure you use free for each malloc you do.
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.
I have written a function that is supposed to read the number of sentences in a .txt file, but when the function is called and done, it gives me a value of 0.
This program over all has 3 more functions to figure out different properties of the file and I have them working great. This one is laid out the same way I wrote my other functions just looking for some advice on why I am getting 0 as my number of sentences.
void ptrCntS (FILE* sp1, int sCount)
{
char sentence = 'O';
int myChr;
if (!(sp1 = fopen("Hello.txt", "r")))
{
printf("error opening Hello.txt");
return(1);
}
while ((myChr = fgetc(sp1)) != EOF)
{
if ('.')
{
sentence ='O';
}
else if (sentence == 'O')
{
sCount++;
sentence = 'I';
}
}
fclose(sp1);
printf ("Total number of sentences are:\t%d", sCount);
return;
}
instead of return use return(sCount);
and assign the return value to some int variable in calling function
like
int sentCount;
.
.
.
sentCount=ptrCntS (param1,param2);
if ('.') is always true, thus else... code never reached. Use if( myChr == '.' ) instead.
Function compiles now and runs properly. This function is being called from a switch in a previous function where I had my addresses set and included my print statement for the totals so that I would not have to write another function in the end to call on all my counts and print their results. Instead I set my case 'A': to call all of my counting functions(in this case that is what the original code is) and than display my results. I am sorry for any lengthiness or my hard to understand writing I am new to the C language and I am having a hard time grasping the literature but making some process on understanding the syntax.
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;
}
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.