I am looking at the code on page 11 here http://www.cs.usfca.edu/~parrt/doc/java/JavaIO-notes.pdf
I have trouble with one statement. I thought the result of an assignment was an lvalue. So ((byteRead = inFile.read()) != -1) should be the same as (inFile.read()) != -1). This doesn't seem to be the case though looking at the output. So my question is how is the statement ((byteRead = inFile.read()) != -1) parsed?
EDIT: It seems from the responses that I had the current interpretation of the result of an assignment. I was wondering what goes wrong by replacing the code fragment
int byteRead;
while((byteRead = inFile.read()) != -1)
outFile.write(byteRead);
with
while( inFile.read() != -1)
outFile.write( inFile.read());
So, now that you posted both versions of code, the answer is clear:
In your first version, each byte read is assigned to byteRead and then written to the output stream.
In the second version, you consume a byte with the read() but don't assign it to a variable. Then, you read another byte (the next one in the stream) which you write to the output stream.
So, if the input file is:
abcdefghijklmnopqrstuvwxyz
The output of the first version will be :
abcdefghijklmnopqrstuvwxyz
The output of the second will be :
bdfhjlnqrtuxz
((byteRead = inFile.read()) != -1) and (inFile.read() != -1) are, in one sense, equivalent boolean expressions. However, the first one has a side effect: It stores the result of inFile.read() in the variable byteRead.
The code example you referenced uses this for a compact while loop that reads one byte from input, writes it to output and keeps doing that until inFile.read() returns -1 (meaning end of file has been reached).
Related
my goal is change a function to a format where the return value of the function is treated :
For example ; treating a the function scanf()
Return value of scanf : The value EOF is returned if the end of input is reached before
either the first successful conversion or a matching failure occurs.
EOF is also returned if a read error occurs, in which case the error
indicator for the stream (see ferror(3)) is set, and errno is set to
indicate the error.
Thus
scanf("%d\n",&i);
will be change to
#define RVAL(exp) do {if ((exp) == -1) { perror (#exp); exit(1); }} while (0)
...
RVAL(scanf("%d\n",&i));
Thus I want this to be done quickly means :
so what i do is look for occurences of "scanf" and replace it with "RVAL(scanf"
but the problem is i have to add another right parentheses
Can this be done fast ? using a techniques ? or a style ? where each whenever I enter scanf(); its replaced witch rval(scanf());
If you don't have many ) in your format string you can use a regex with (scanf([^)]*)); and replace with rval(\1);
*see comment
I was working on OVM driver sequencer communication. I am using try_get_item() in ovm driver but it is still getting stuck. In my sequencer I redefined try_next_item and just printed a display statement before and after m_req_fifo.peek(t); The statement before peek got executed but not the statement after the peek. I even displayed size of the m_req_fifo using m_req_fifo.size() and it printed out 1. Why is peek not returning anything even after the size is 1? The modified try_next_item (Just addition of display) is given below.
The line After PEEK never gets executed after the line Line 398 with fifo size 1
virtual task try_next_item(output REQ t);
int selected_sequence;
time arb_time;
ovm_sequence_base seq;
if (get_next_item_called == 1) begin
ovm_report_error(get_full_name(), "get_next_item/try_next_item called twice without item_done or get in between", OVM_NONE);
return;
end
wait_for_sequences();
selected_sequence = choose_next_request();
if (selected_sequence == -1) begin
t = null;
return;
end
set_arbitration_completed(arb_sequence_q[selected_sequence].request_id);
seq = arb_sequence_q[selected_sequence].sequence_ptr;
arb_sequence_q.delete(selected_sequence);
m_update_lists();
sequence_item_requested = 1;
get_next_item_called = 1;
$display("Line 398 with fifo size %0d\n", m_req_fifo.size());
m_req_fifo.peek(t);
$display("After PEEK\n");
wait_for_sequences();
// attempt to get the item; if it fails, produce an error and return
if (!m_req_fifo.try_peek(t))
ovm_report_error("TRY_NEXT_BLOCKED", {"try_next_item: the selected sequence '",
seq.get_full_name(), "' did not produce an item during wait_for_sequences(). ",
"Sequences should not consume time between calls to start_item and finish_item. ",
"Returning null item."}, OVM_NONE);
endtask
uvm_tlm_fifo::size() doesn't return the number of elements in the FIFO, but its capacity (i.e. the maximum number of elements it can hold). The function you're looking for is uvm_tlm_fifo::used() which returns the number of stored elements.
The function names are not intuitive at all and I remember spending a couple of hourse trying to understand some similar code to the one you had until noticing in the documentation that I was using the wrong method.
I have a mex function that takes in a field of a struct in the third input (i.e. prhs[2]), which is a boolean. If true, it will parse information from the fourth input (i.e. prhs[3]). In a nutshell, this is the code excerpt:
mxValue = mxGetField(prhs[3], 0, "change"); mxLogical *change;
change = mxGetLogicals(mxValue);
mexPrintf("true/false: %i \n", *change);
mexEvalString("drawnow;");
if ( change ) {
mexPrintf("...Parsing info... \n");
mexEvalString("drawnow;");
mxValue = mxGetField(prhs[3], 0, "info");
nRows = mxGetM(mxValue); nCols = mxGetN(mxValue);
Eigen::Map<Eigen::VectorXd> info((double *)mxGetPr(mxValue),nRows);
}
As you can see, I do a printout to see whether the input prhs[2] is true or false. Even if the function prints out false, the if statement gets executed regardless, because I can see the printout ...Parsing info....
Why is my MATLAB mex function ignoring my if statement?
C is not MATLAB! C is C!
You are checking if pointer change has a value. It does indeed have a value, a memory direction e.g. #72BA21, to the location where the value of the boolean is stored.
You can either check the contents of whats inside that specific direction if(*change) as #buzjwa suggest, or grab the information on the array, instead of a pointer to it, using mxGetData.
As a side note: learn to debug, or at least, print statements. a simple mexPrintf() call would have shown you what change contains
I have tried to get multiple user inputs to print them in Scala IDE.
I have tried the this piece of code
println(scala.io.StdIn.readLine())
which works, as the IDE takes my input and then print it in the line but this works only for a single input.
I want the code to take multiple inputs till only newline is entered. example,
1
2
3
so i decided we needed an iterator for the input, which led me to try the following 2 lines of code seperately
var in = Iterator.continually{ scala.io.StdIn.readLine() }.takeWhile { x => x != null}
and
var in = io.Source.stdin.getLines().takeWhile { x => x != null}
Unfortunately none of them worked as the IDE is not taking my input at all.
You're really close.
val in = Iterator.continually(io.StdIn.readLine).takeWhile(_.nonEmpty).toList
This will read input until an empty string is entered and saves the input in a List[String]. The reason for toList is because an Iterator element doesn't become real until next is called on it, so readLine won't be called until the next element is required. The transition to List creates all the elements of the Iterator.
update
As #vossad01 has pointed out, this can be made safer for unexpected input.
val in = Iterator.continually(io.StdIn.readLine)
.takeWhile(Option(_).fold(false)(_.nonEmpty))
.toList
I'm java programmer, but I used to use C++ long time ago. Now I have to translate form C++ to Java one program and I found one line, wchich I don't know what does it mean:
if (wr[m-2] == 0) wr[m-1] == 0;
Program works correctly in C++, but of course there's an AssignmentOperator error in Eclipse.
Looks like you're checking
Based on wr being an array of ints and m being an int, if the array wr is equal to 0 at index m-2, then check if the wr at index m-1 is equal to 0.
That line looks right, but you might want to check that your minus signs are actually minus signs and not a different-but-similar type of hyphen. Otherwise it looks fine to me.
The lines you quoted result in a no-op where there likely should be an assignment.
if (wr[m-2] == 0)
wr[m-1] == 0;
should probably be
if (wr[m-2] == 0)
wr[m-1] = 0;