Hi, I'm a lil new to this C stuff, but I am having some trouble finding a way to count the number of times the "on" characters appear in a paragraph.The paragraph comes from a seperate input file and has to be sent out to an output file with a summary of information about the paragraph, one thing being the number of times the "on" characters appeared. Any help is greatly appreciated.
Post what you've tried so far and someone will help you.
Basically:
- read a character
- if it's one you're interested in, increment counter
- repeat until end
When all else fails, read the instructions.
If you're posting code, use code tags: [code] /* insert code here */ [/code]
okay,
This is where I get stuck, b/c it has to ignore whitespace and new lines. Yet it must check for the 'n' also. Thanks again, as always any help is greatly appreciated!Code:#include <stdio.h> #include <ctype.h> int main(void) { in c; while ((c = getchar()) != EOF) if (c='o'))
"on" is not a character, it is a string, might want to try strstr to look for it, and when comparing characters for equality like you did in your previous post (or other variables for that matter), you use ==, not = which is the assignment operator.
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i'm not familiar with strstr..
and == comment is noted.. thank you
if anyone could help me with strstr or another method... I would appreciate it
mabey you should use strtok, I haven't used it much, it could be done with strstr like I stated earlier, but trying to write the code came out clunky and inefficient.
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okay, i have another quick question.. I'm making a word count program too and it isn't giving me the output for some reason.. i've been up for like 35 hours so it is astonishingly hard to read code right now.. but again thanks in advance
Code:// the following code counts the number of words #include <stdio.h> #include <ctype.h> int found_next_word(void); int main (void) { int word_count=0; while(found_next_word()==1) ++word_count; printf("Number of words = %d\n\n", word_count); return 0; } int found_next_word(void) { int c; while (isspace(c = getchar())) ; /*skip white space*/ if (c != EOF) { /*found a word*/ while ((c = getchar()) != EOF && !isspace(c)) ; /*skip all but EOF and white space*/ return 1; } return 0; }
Here are some examples that do *almost* what you're trying to do.
Using strstr:
And using a more 'manual' approach.Code:int count(const char * buff, const char * what) { int cnt = 0; while((buff=strstr(buff,what))!=0) { ++cnt, buff++; } return cnt; }
Code:int count(const char * buff, const char * what) { int cnt = 0; int good = 0; int i = 0; int j = 0; while(buff[i]!=0) { if(good==0) { //reset j=0; } else if(what[j]==0) { //got it ++cnt; j=0; } if(buff[i]==what[j]) { //so far, so good good=1; } else { good=0; } ++i, j++; } if(what[j]==0) { ++cnt; } return cnt; }
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
Okay, here's what I am ultimatly trying to do:
Given an input paragraph find the number of words, lines, and number of times the character combination "on" appears. Can anyone here help me combine some of this?
Slightly confused. Based on what you said here:
Are we to assume in this paragraph:Originally posted by miltecnico
...it has to ignore whitespace and new lines. Yet it must check for the 'n' also.
there are 6, 4, or 2 "on"s?Only on the way to nowhere can the two
new members reach one answer on Tuesday
For this part of your problem and using the code you posted first, if you find an "o" simply read the next character and process it accordingly within the if
Definition: Politics -- Latin, from
poly meaning many and
tics meaning blood sucking parasites
-- Tom Smothers
Yes, there are 6 on's in that paragraph. I know the theory behind it already but I don't know the code...
Given "sno now"
I need something that would read in 's', 'n', 'o'...
trigger a loop to then check if it's followed by an 'n'
and up a counter if yes
or skip if no,
and continue to read each letter... 'o','w' until EOF
regarding new lines.. Given an input paragraph it has to read in each character and the program will eventually cipher the paragraph using a different alphabet and when a "'/n'"character is read in it will be replaced with the same "'/n'" character
You could always do it the easy way...
Code:while( (c=fgetc( fp )) != EOF ) { if( c == 'o' ) { if( (c = fgetc( fp )) == 'n' ) count++; else ungetc( c, fp ); } }
Quzah.
Hope is the first step on the road to disappointment.
would you be willing to explain that a little, I'm kind of new to this still?
