I am writing a program where I am printing the prime multiples of a given number and am using a print function like this
printf("%d*", variable);
I was hoping that it would return something like
"2*2*2*3" but, naturally, at the end there is an extra asterisk "2*2*2*3*"
I was wondering if there was a quick fix to this or something else I can try. Thanks
Change it LIke this
printf("%d",variable); //for first number
printf("*%d",variable); //use this for printing other numbers
Related
I have a lot of excel files looking like that:
Example:
My goal is to make it look like that:
To do that, I used very simple excel's function:
=F7&" "&G7&".........cat."&" "&H7&" times "&I7&CHAR(10)&F8&" "&G8&".........cat."&" "&H8&" times "&I8&CHAR(10)
The thing is, the number of dots placed before "cat" is not constant. It depends where the previous sentence ends and my formula doesn't take it into account - it always adds 9 dots, which means I have to add the rest of the dots manually.
Any ideas how to make it work? :D
The REPT function can do this. Use LEN to calculate the length of what you're adding the dots to, then subtract that from the desired width of the result. That will repeat the dot enough times to fill the column. For example, if you want the text with dots to be 40 characters, right padded with .:
=F1&" "&G1&REPT(".",40-LEN(G1))&"cat."&" "&H1&" times "&I1&CHAR(10)&F2&""
=LEFT(A1 & REPT(".",22-LEN(A1))&"cat",25)
22 = fixed width - len("cat"), 25 - fixed width.
edit - i revised because my original answer was not correct but I see Comintern has posted a similar response since.
In my mind this should be easy.. I have spent a good bit of time trying to get this right
Problem-
I have 1 data set that returns whole numbers as well as percents. What I am looking for is a formatting step to work and add the correct suffix (x100+% when % or nothing)
Here is what I have but don't get consistent results
=iif(Fields!Mid_Size.Value<1,Format(Fields!Mid_Size.Value,"P"),Format(Fields!Mid_Size.Value,"#"))
The raw data looks like:
Alpha Mid-Size
11 49
0.0718954248366013 0.320261437908497
Anyone have any ideas?
Try this:
=iif(Fields!Mid_Size.Value<1,FormatPercent(Fields!Mid_Size.Value,0),Format(Fields!Mid_Size.Value,"#"))
This uses the FormatPercent function. The '0' is for no decimal places; you can set that to however many you want.
i am having a little trouble with printf specifiers...so before asking you guys i read almost everything onC++Reference page, but couldnt fix the problem, and since i am new at c i cant even understand the problem, its most likely a syntax error but i can't find it...
for(i = 1; i <= 10; i++) {
printf("\n%d.%s%n",i,names[i-1],offset);
printf("%*s%.2f TL",10-offset," ",prices[i-1]);
}
so basically i have this code to print a list, and i want the prices to start from the same column.
For e.g:
water 1.00
oj 1.00
and the logic behind my code (incase it's not obvious, i can't tell if it is) is:
print id number and name, count how many chars we've written so far and assign it to offset.
print (starting column of price list)-offset spaces before price
once i couldn't get the result i want, i checked and found out that offset is 3 for all names which is not the case(and no value is assigned to offset before this procedure).
Thanks for any kind of help !
PS: This is a practice code just to get better at using specifiers efficiently.
edit:
so i did this :
for(i=1;i<=10;i++)
{
printf("%d.%s%n",i,names[i-1],&offset);
printf("%*s%.2f TL\n",10-offset," ",prices[i-1]);
}
but what i get as a result is huge empty black command screen.
The %n format specifier requires a pointer. Your code is missing the & operator for offset:
printf("\n%d.%s%n",i,names[i-1],&offset);
The good ol' C interface doesn't know what types you supply to printf so it doesn't complain and happily reads the 4 byte integer value of offset on the stack as a memory location -> core dump.
Actually, g++ with -Wall does warn. So
hd1 has a point here because C++ output is type safe (even though it's a pain);
Heed thy warnings.
When you use %n in a printf format, the corresponding parameter must be a pointer. printf will store the information in the int you point it to.
Assuming you've declared int offset somewhere, you should use &offset as the last argument in your printf call.
While we're here, allow me to comment on this excerpt:
printf("\n
ARGH NO! Newline is a terminator. It goes at the end of a line, not the beginning.
so i did this :
for(i=1;i<=10;i++)
{
printf("%d.%s%n",i,names[i-1],&offset);
printf("%*s%.2f TL\n",10-offset," ",prices[i-1]);
}
but what i get as a result is huge empty black command screen.
edit: Can you guys try this and tell me if you get normal results? I can't understand the mistake occuring, so i can't get past it...Maybe some other examples will lead me to where is the mistake.
I have some decimal numbers that I need to write to a text file with leading zeros when appropriate. I've done some research on this, and everything I've seen suggests something like:
REAL VALUE
INTEGER IVALUE
IF (VALUE.LT.0) THEN
IVALUE = CEILING(VALUE)
ELSE
IVALUE = FLOOR(VALUE)
ENDIF
WRITE(*,1) IVALUE, ABS(VALUE)-ABS(IVALUE)
1 FORMAT(I3.3,F5.4)
As I understand it, the IF block and ABS parts should allow this to work for all values on -100 < VALUE < 1000. If I set VALUE = 12.3456, the code above should produce "012.3456" as the output, and it does. However if I have something like VALUE = -12.3456, I'm getting "(3 asterisks).3456" as my output. I know the asterisks usually shows up when there are not enough characters provided for in the FORMAT statement, but 3 should be enough in this example (1 character for the "-" and two characters for "12"). I haven't tested this yet with something like VALUE = -9.876, but I'd expect the output to be "-09.8760".
Is there something wrong in my understanding of how this works? Or is there some other limitation of this technique that I'm violating?
UPDATE: Okay I've looked into this some more, and it seems to be a combination of a negative value and the I3.3 format. If VALUE is positive and I have the I3.3, it will put leading zeros as expected. If VALUE is negative and I only have I3 as my format, I get the correct value output, but it will be padded with spaces before the negative sign instead of padded with zeros after the negative (so -9.8765 is output as " -9.8765", but that leading space breaks what I'm using the .txt file for, so it's not acceptable).
Tho problem is with your integer data edit descriptor. With I3.3 you require at least 3 digits and the field width is only 3. There is no place for the minus sign. Use I4.3 or, In Fortran 95 and above, I0.3.
Answer to your edit: Use I0.3, it uses the minimum number of characters necessary.
But finally, you just probably want this: WRITE(*,'(f0.3)') VALUE
Of course, I could get what I'm looking for by changing it up a little bit to
REAL VALUE
INTEGER IVALUE
IF (VALUE.LT.0) THEN
WRITE(*,1) FLOOR(ABS(IVALUE)), ABS(VALUE)-FLOOR(ABS(VALUE))
1 FORMAT('-',I2.2,F5.4)
ELSE
WRITE(*,2) FLOOR(VALUE), ABS(VALUE)-FLOOR(BS(VALUE))
2 FORMAT(I3.3,F5.4)
ENDIF
But this feels a lot clunkier, and in reality I'm going to try to be writing multiple values in the same line, which will lead to really messy IF blocks or complex cursor movement, which I'd like to avoid if at all possible.
as another way to skin the cat.. I'd prefer not to do arithmatic on the data at all but just work on the format:
character*8 fstring/'(f000.4)'/
val=12.34
if(val.gt.1)then
write(fstring(3:5),'(i0)')6+floor(log10(val))
elseif(val.lt.-1)then
write(fstring(3:5),'(i0)')7+floor(log10(-val))
elseif(val.ge.0)
write(fstring(3:5),'(i0)')6
else
write(fstring(3:5),'(i0)')7
endif
write(*,fstring)val
just for fun with modern fortran that supports character functions you can roll that up in a function and end up with a construct like this:
write(*,'('//fstring(val1)//','//fstring(val2)//')')val1,val2
having a bit of trouble finding a solution to this.
I want to take a large ordered text file of words and create - in the same order - a text file of fixed length numeric values.
For example:
Input File Output File
AAA -> 00000001
AAH -> 00002718
AAZ -> 71827651
Initially it seemed a hash function would do the trick. However they are one way. Also perhaps they are a bit "heavyweight" for this. After all, I don't need any cryptography. Plus, it's a reference file. It will never change.
Any compression is a bonus not essential. That said, I don't want the file to get any bigger than it already is. Which is why I don't just want to write out the words as text but with fixed lengths.
So, bottom line; input is a NSString of variable length, output is an integer of fixed length. And, I must be able to take the integer and figure out the string.
Any help much appreciated!
Thanks!
xj
Well, this would be a bit of a brute force method, but here's my guess.
Start by making a custom function to convert one letter of text to an integer less than 100. (I'm not sure if such a function already exists, if so then great!) You might need to just go to stuff like "if ([input isEqual: #"a"]){ return 1;}
Then, run that function on each letter of text, and get the final integer by combining the previous results.
For example:
int myVal1 = [intConverter firstLetter];
int myVal2 = [intConverter secondLetter];
int myVal3 = [intConverter thirdLetter];
int finalValue =100^3 + 100^2*myVal1 + 100*myVal2 + myVal3;
Then, finalValue would be of the form 1(myVal1)(myVal2)(myVal3), which is what I think you're looking for.
To get back the original string, simply use the mod (%) and division functions to get the individual values back, then run the intConverter function backwards. (This would probably mean writing a new function that basically runs those if statements in reverse, but oh well.)
I hope this helps.