int count_words(string word)
{
string spaces = "";
int total_words = 1;
int i, j = 0 ;
for (i = 0; i < strlen(word); i++)
{
strcpy(spaces, word[i]);
if (strcmp(spaces, " ") == 0)
{
total_words = total_words + 1;
}
}
return total_words;
}
I am trying to make a function in c that gets the total number of words, and my strategy is to find the number of spaces in the string input. However i get an error at strcpy about integer to ptrtype conversion,. I cant seem to compare the 2 strings without getting the error. Can someone explain to me whats how the error is happening and how I would go about fixing it. The IDE is also suggesting me to add an ampersand beside word[i] but it then makes a segmentation fault output
You need to learn a little more about the distinction between characters and strings in C.
When you say
strcpy(spaces, word[i]);
it looks like you're trying to copy one character to a string, so that in the next line you can do
if (strcmp(spaces, " ") == 0)
to compare the string against a string consisting of one space character.
Now, it's true, if you're trying to compare two strings, you do have to call strcmp. Something like
if (spaces == " ") /* WRONG */
definitely won't cut it.
In this case, though, you don't need to compare strings. You're inspecting your input a character at a time, so you can get away with a simple character comparison instead. Get rid of the spaces string and the call to strcpy, and just do
if (word[i] == ' ')
Notice that I'm comparing against the character ' ', not the string " ". Using == to compare single characters like this is perfectly fine.
Sometimes, you do have to construct a string out of individual characters, "by hand", but when you do, it's a little more elaborate. It would look like this:
char spaces[2];
spaces[0] = word[i];
spaces[1] = '\0';
if (strcmp(spaces, " ") == 0)
...
This would work, and you might want to try it to be sure, but it's overkill, and there's no reason to write it that way, except perhaps as a learning exercise.
Why didn't your code
strcpy(spaces, word[i]);
work? What did the error about "integer to pointer conversion" mean? Actually there are several things wrong here.
It's not clear what the actual type of the string spaces is (more on this later), but it has space for at most 0 characters, so you're not going to be able to copy a 1-character string into it.
It's also not clear that spaces is even writable. It might be a constant, meaning that you can't legally copy any characters into it.
Finally, strcpy copies one string to another. In C, although strings are arrays, they're usually referred to as pointers. So strcpy accepts two pointers, one to the source and one to the destination string. But you passed it word[i], which is a single character, not a string. Let's say the character was A. If you hadn't gotten the error, and if strcpy had tried to do its job, it would have treated A as a pointer, and it would have tried to copy a string from address 65 in memory (because the ASCII value of the character A is 65).
This example shows that working with strings is a little bit tricky in C. Strings are represented as arrays, but arrays are second-class citizens in C. You can't pass arrays around, but arrays are usually referred to by simple pointers to their first element, which you can pass around. This turns out to be very convenient and efficient once you understand it, but it's confusing at first, and takes some getting used to.
It might be nice if C did have a built-in, first-class string type, but it does not. Since it does not, C programmers myst always keep in mind the distinction between arrays and characters when working with strings. That "convenient" string typedef they give you in CS50 turns out to be a bad idea, because it's not actually convenient at all -- it merely hides an important distinction, and ends up making things even more confusing.
Related
I was trying to remove spaces of a string after scanning it. The program has compiled perfectly but it is not showing any output and the output screen just keeps getting shut down after scanning the string. Is there a logical error or is there some problem with my compiler(I am using a devc++ compiler btw).
Any kind of help would be appreciated
int main()
{
char str1[100];
scanf("%s",&str1);
int len = strlen(str1);
int m;
for (m=0;m<=len;){
if (&str1[m]==" "){
m++;
}
else {
printf("%c",&str1[m]);
}
m++;
}
return 0;
}
Edit : sorry for the error of m=1, I was just checking in my compiler whether that works or not and just happened to paste that code
Your code contains a lot of issues, and the behaviour you describe is very likely not because of a bug in the compiler :-)
Some of the issues:
Use scanf("%s",str1) instead of scanf("%s",&str1). Since str1 is defined as a character array, it automatically decays to a pointer to char as required.
Note that scanf("%s",str1) will never read in any white space because "%s" is defined as skipping leading white spaces and stops reading in when detecting the first white space.
In for (m=1;m<=len;) together with str1[m], note that an array in C uses zero-based indizes, i.e. str1[0]..str1[len-1], such that m <= len exceeds array bounds. Use for (m=0;m<len;) instead.
Expression &str1[m]==" " is correct from a type perspective, but semantically a nonsense. You are comparing the memory address of the mth character with the memory address of a string literal " ". Use str1[m]==' ' instead and note the single quotes denoting a character value rather than double quotes denoting a string literal.
Statement printf("%c",&str1[m]) passes the memory address of a character to printf rather than the expected character value itself. Use printf("%c",str1[m]) instead.
Hope I found everything. Correct these things, turn on compiler warnings, and try to get ahead. In case you face further troubles, don't hesitate to ask again.
Hope it helps a bit and good luck in experiencing C language :-)
There are many issues:
You cannot read a string with spaces using scanf("%s") , use fgets instead (see comments).
scanf("%s", &str1) is wrong anyway, it should be scanf("%s", str1);, str1 being already the address of of the string
for (m = 0; m <= len;) is wrong, it should be for (m = 0; m < len;), because otherwise the last character you will check is the NUL string terminator.
if (&str1[m]==" ") is wrong, you should write if (str1[m]==' '), " " does not denote a space character but a string literal, you need ' ' instead..
printf("%c", &str1[m]); is wrong, you want to print a char so you need str1[m] (without the &).
You should remove both m++ and put that into the for statement: for (m = 1; m < len; m++), that makes the code clearer.
And possibly a few more problems.
And BTW your attempt doesn't remove the spaces from the string, it merely displays the string skipping spaces.
There are a number of smaller errors here that are adding up.
First, check the bounds on your for loop. You're iterating from index 1 to index strlen(str1), inclusive. That's a reasonable thing to try, but remember that in C, string indices start from 0 and go up to strlen(str1), inclusive. How might you adjust the loop to handle this?
Second, take a look at this line:
if (&str1[m] == " ") {
Here, you're attempting to check whether the given character is a space character. However, this doesn't do what you think it does. The left-hand side of this expression, &str1[m], is the memory address of the mth character of the string str1. That should make you pause for a second, since if you want to compare the contents of memory at a given location, you shouldn't be looking at the address of that memory location. In fact, the true meaning of this line of code is "if the address of character m in the array is equal to the address of a string literal containing the empty string, then ...," which isn't what you want.
I suspect you may have started off by writing out this line first:
if (str1[m] == " ") {
This is closer to what you want, but still not quite right. Here, the left-hand side of the expression correctly says "the mth character of str1," and its type is char. The right-hand side, however, is a string literal. In C, there's a difference between a character (a single glyph) and a string (a sequence of characters), so this comparison isn't allowed. To fix this, change the line to read
if (str1[m] == ' ') {
Here, by using single quotes rather than double quotes, the right-hand side is treated as "the space character" rather than "a string containing a space." The types now match.
There are some other details about this code that need some cleanup. For instance, look at how you're printing out each character. Is that the right way to use printf with a character? Think about the if statement we discussed above and see if you can tinker with your code. Similarly, look at how you're reading a string. And there may be even more little issues here and there, but most of them are variations on these existing themes.
But I hope this helps get you in the right direction!
For loop should start from 0 and less than length (not less or equal)
String compare is wrong. Should be char compare to ' ' , no &
Finding apace should not do anything, non space outputs. You ++m twice.
& on %c output is address not value
From memory, scanf stops on whitespace anyway so needs fgets
int main()
{
char str1[100];
scanf("%s",str1);
int len = strlen(str1);
int m;
for (m=0;m<len;++m){
if (str1[m]!=' '){
printf("%c",str1[m]);
}
}
return 0;
}
There are few mistakes in your logic.
scanf terminates a string when it encounter any space or new line.
Read more about it here. So use fgets as said by others.
& in C represents address. Since array is implemented as pointers in C, its not advised to use & while getting string from stdin. But scanf can be used like scanf("%s",str1) or scanf("%s",&str[1])
While incrementing your index put m++ inside else condition.
Array indexing in C starts from 0 not 1.
So after these changes code will becames something like
int main()
{
char str1[100];
fgets(str1, sizeof str1 , stdin);
int len = strlen(str1);
int m=0;
while(m < len){
if (str1[m] == ' '){
m++;
}
else {
printf("%c",str1[m]);
m++;
}
}
return 0;
}
in one of my university assignments I am restricted in the libraries I use. I am new to C and pointers and want to see if two strings (or should I say char's) are equal.
Part of me wants to loop through every char of the 'char string' and test equivalence, but then it comes back how to test equivalence (lol).
Any help is appreciated.
edit: I am seeing this:
warning: result of comparison against a string literal is
unspecified (use strncmp instead) [-Wstring-compare]
which leads to a segmentation fault. I know it has to do with this piece of code because all I added was:
if (example.name == "testName"){
printf("here!\n");
}
Part of me wants to loop through every char of the 'char string' and test equivalence
That's exactly what you need to do. Make a function mystrcmp with the signature identical to regular strcmp,
int mystrcmp ( const char * str1, const char * str2 );
and write your own implementation.
but then it comes back how to test equivalence.
When you loop character-by-character, you test equivalence of individual characters, not strings. Characters in C can be treated like numbers: you can compare them for equality using ==, check what character code is less than or greater than using < and >, and so on.
The only thing left to do now is deciding when to stop. You do that by comparing the current character of each string to zero, which is the null terminator.
Don't forget to forward-declare your mystrcmp function before using it.
A string in C is terminated with null character(0x00 or \0).You should compare both strings in a loop character by character till null char for either of the string is reached.
Loop should be broken if characters are not equal.
EDIT:
To answer your edit in question:
You should take two character pointers pointing to both strings and then copmare them like
//loop start,loop till null for any one of the string is found
if(*ptr1 != *ptr2)
{
//break loop
}
ptr1++;ptr2++;
//end loop
if((*ptr1 == *ptr2) &&(*ptr1== 0x00))
{
//strings are equal
}
Given that this is a university assignment, you should pay heed to chars just being small integers. You should also pay heed that C strings are contiguous memory buffers terminated by a binary zero (0x00).
You should also learn about pointer math. You will learn ways to shorten the code you have to write while learning something really interesting concerning the C language and how computers work. It will certainly help you if you choose a career on lower-level programming.
I am originally a Java programmer who is now struggling with C and specifically C's pointers.
The idea on my mind is to receive a string, from the user, on a command line, into a character pointer. I then want to access its individual elements. The idea is later to devise a function that will reverse the elements' order. (I want to work with anagrams in texts.)
My code is
#include <stdio.h>
char *string;
int main(void)
{
printf("Enter a string: ");
scanf("%s\n",string);
putchar(*string);
int i;
for (i=0; i<3;i++)
{
string--;
}
putchar(*string);
}
(Sorry, Code marking doesn't work).
What I am trying to do is to have a first shot at accessing individual elements. If the string is "Santillana" and the pointer is set at the very beginning (after scanf()), the content *string ought to be an S. If unbeknownst to me the pointer should happen to be set at the '\0' after scanf(), backing up a few steps (string-- repeated) ought to produce something in the way of a character with *string. Both these putchar()'s, though, produce a Segmentation fault.
I am doing something fundamentally wrong and something fundamental has escaped me. I would be eternally grateful for any advice about my shortcomings, most of all of any tips of books/resources where these particular problems are illuminated. Two thick C books and the reference manual have proved useless as far as this.
You haven't allocated space for the string. You'll need something like:
char string[1024];
You also should not be decrementing the variable string. If it is an array, you can't do that.
You could simply do:
putchar(string[i]);
Or you can use a pointer (to the proposed array):
char *str = string;
for (i = 0; i < 3; i++)
str++;
putchar(*str);
But you could shorten that loop to:
str += 3;
or simply write:
putchar(*(str+3));
Etc.
You should check that scanf() is successful. You should limit the size of the input string to avoid buffer (stack) overflows:
if (scanf("%1023s", string) != 1)
...something went wrong — probably EOF without any data...
Note that %s skips leading white space, and then reads characters up to the next white space (a simple definition of 'word'). Adding the newline to the format string makes little difference. You could consider "%1023[^\n]\n" instead; that looks for up to 1023 non-newlines followed by a newline.
You should start off avoiding global variables. Sometimes, they're necessary, but not in this example.
On a side note, using scanf(3) is bad practice. You may want to look into fgets(3) or similar functions that avoid common pitfalls that are associated with scanf(3).
This might seem like a very simple question, but I am struggling with it. I have been writing iPhone apps with Objective C for a few months now, but decided to learn C Programming to give myself a better grounding.
In Objective-C if I had a UILabel called 'label1' which contained some text, and I wanted to run some instructions based on that text then it might be something like;
if (label1.text == #"Hello, World!")
{
NSLog(#"This statement is true");
}
else {
NSLog(#"Uh Oh, an error has occurred");
}
I have written a VERY simple C Program I have written which uses printf() to ask for some input then uses scanf() to accept some input from the user, so something like this;
int main()
{
char[3] decision;
Printf("Hi, welcome to the introduction program. Are you ready to answer some questions? (Answer yes or no)");
scanf("%s", &decision);
}
What I wanted to do is apply an if statement to say if the user entered yes then continue with more questions, else print out a line of text saying thanks.
After using the scanf() function I am capturing the users input and assigning it to the variable 'decision' so that should now equal yes or no. So I assumed I could do something like this;
if (decision == yes)
{
printf("Ok, let's continue with the questions");
}
else
{
printf("Ok, thank you for your time. Have a nice day.");
}
That brings up an error of "use of undeclared identifier yes". I have also tried;
if (decision == "yes")
Which brings up "result of comparison against a string literal is unspecified"
I have tried seeing if it works by counting the number of characters so have put;
if (decision > 3)
But get "Ordered comparison between pointer and integer 'Char and int'"
And I have also tried this to check the size of the variable, if it is greater than 2 characters it must be a yes;
if (sizeof (decision > 2))
I appreciate this is probably something simple or trivial I am overlooking but any help would be great, thanks.
Daniel Haviv's answer told you what you should do. I wanted to explain why the things you tried didn't work:
if (decision == yes)
There is no identifier 'yes', so this isn't legal.
if (decision == "yes")
Here, "yes" is a string literal which evaluates to a pointer to its first character. This compares 'decision' to a pointer for equivalence. If it were legal, it would be true if they both pointed to the same place, which is not what you want. In fact, if you do this:
if ("yes" == "yes")
The behavior is undefined. They will both point to the same place if the implementation collapses identical string literals to the same memory location, which it may or may not do. So that's definitely not what you want.
if (sizeof (decision > 2))
I assume you meant:
if( sizeof(decision) > 2 )
The 'sizeof' operator evaluates at compile time, not run time. And it's independent of what's stored. The sizeof decision is 3 because you defined it to hold three characters. So this doesn't test anything useful.
As mentioned in the other answer, C has the 'strcmp' operator to compare two strings. You could also write your own code to compare them character by character if you wanted to. C++ has much better ways to do this, including string classes.
Here's an example of how you might do that:
int StringCompare(const char *s1, const char *s2)
{ // returns 0 if the strings are equivalent, 1 if they're not
while( (*s1!=0) && (*s2!=0) )
{ // loop until either string runs out
if(*s1!=*s2) return 1; // check if they match
s1++; // skip to next character
s2++;
}
if( (*s1==0) && (*s2==0) ) // did both strings run out at the same length?
return 0;
return 1; // one is longer than the other
}
You should use strcmp:
if(strcmp(decision, "yes") == 0)
{
/* ... */
}
You should be especially careful with null-terminated string in C programming. It is not object. It is a pointer to a memory address. So you can't compare content of decision directly with a constant string "yes" which is at another address. Use strcmp() instead.
And be careful that "yes" is actually "yes\0" which will take 4 bytes and the "\0" is very important to strcmp() which will be recognized as the termination during the comparison loop.
Ok a few things:
decision needs to be an array of 4 chars in order to fit the string "yes" in it. That's because in C, the end of a string is indicated by the NUL char ('\0'). So your char array will look like: { 'y', 'e', 's', '\0' }.
Strings are compared using functions such as strcmp, which compare the contents of the string (char array), and not the location/pointer. A return value of 0 indicates that the two strings match.
With: scanf("%s", &decision);, you don't need to use the address-of operator, the label of an array is the address of the start of the array.
You use strlen to get the length of a string, which will just increment a counter until it reaches the NUL char, '\0'. You don't use sizeof to check the length of strings, it's a compile-time operation which will return the value 3 * sizeof(char) for a char[3].
scanf is unsafe to use with strings, you should alternatively use fgets(stdin...), or include a width specifier in the format string (such as "3%s") in order to prevent overflowing your buffer. Note that if you use fgets, take into account it'll store the newline char '\n' if it reads a whole line of text.
To compare you could use strcmp like this:
if(strcmp(decision, "yes") == 0) {
// decision is equal to 'yes'
}
Also you should change char decision[3] into char decision[4] so that the buffer has
room for a terminating null character.
char decision[4] = {0}; // initialize to 0
There's several issues here:
You haven't allocated enough storage for the answer:
char[3] decision;
C strings are bytes in the string followed by an ASCII NUL byte: 0x00, \0. You have only allocated enough space for ye\0 at this point. (Well, scanf(3) will give you yes\0 and place that NUL in unrelated memory. C can be cruel.) Amend that to include space for the terminating \0 and amend your scanf(3) call to prevent the buffer overflow:
char[4] decision;
/* ... */
scanf("%3s", decision);
(I've left off the &, because simply giving the name of the array is the same as giving the address of its first element. It doesn't matter, but I believe this is more idiomatic.)
C strings cannot be compared with ==. Use strcmp(3) or strncmp(3) or strcasecmp(3) or strncasecmp(3) to compare your strings:
if(strcasecmp(decision, "yes") == 0) {
/* yes */
}
C has lots of lib functions to handle this but it pays to know what you are declaring.
Declaring
char[3] decision;
is actually declaring a char array of length 3. So therefor attempting a comparison of
if(decision == "yes")
is comparing a literal against and array and therefor will not work. Since there is no defined string type in C you have to use pointers, but not directly, if you don't want to. In C strings are in fact arrays of char so you can declare them both ways eg:
char[3] decision ;
* char decision ;
Both will in point of fact work but you in the first instance the compiler will allocate the memory for you, but it will ONLY allocate 3 bytes. Now since strings in C are null terminated you need to actually allocate 4 bytes since you need room for "yes" and the null. Declaring it the second way simply declares a pointer to someplace in memory but you have no idea really where. You would then have to allocate memory to contain whatever you are going to put there since to do otherwise will more then likely cause a SEGFAULT.
To compare what you get from input you have two options, either use the strcomp() function or do it yourself by iterating through decision and comparing each individual byte against "Y" and "E" and "S" until you hit null aka \0.
There are variations on strcomp() to deal with uppercase and lowercase and they are part of the standard string.h library.
I'm trying to use Mac OS X's listxattr C function and turn it into something useful in Python. The man page tells me that the function returns a string buffer, which is a "simple NULL-terminated UTF-8 strings and are returned in arbitrary order. No extra padding is provided between names in the buffer."
In my C file, I have it set up correctly it seems (I hope):
char buffer[size];
res = listxattr("/path/to/file", buffer, size, options);
But when I got to print it, I only get the FIRST attribute ONLY, which was two characters long, even though its size is 25. So then I manually set buffer[3] = 'z' and low and behold when I print buffer again I get the first TWO attributes.
I think I understand what is going on. The buffer is a sequence of NULL-terminated strings, and stops printing as soon as it sees a NULL character. But then how am I supposed to unpack the entire sequence into ALL of the attributes?
I'm new to C and using it to figure out the mechanics of extending Python with C, and ran into this doozy.
char *p = buffer;
get the length with strlen(p). If the length is 0, stop.
process the first chunk.
p = p + length + 1;
back to step 2.
So you guessed pretty much right.
The listxattr function returns a bunch of null-terminated strings packed in next to each other. Since strings (and arrays) in C are just blobs of memory, they don't carry around any extra information with them (such as their length). The convention in C is to use a null character ('\0') to represent the end of a string.
Here's one way to traverse the list, in this case changing it to a comma-separated list.
int i = 0;
for (; i < res; i++)
if (buffer[i] == '\0' && i != res -1) //we're in between strings
buffer[i] = ',';
Of course, you'll want to make these into Python strings rather than just substituting in commas, but that should give you enough to get started.
It looks like listxattr returns the size of the buffer it has filled, so you can use that to help you. Here's an idea:
for(int i=0; i<res-1; i++)
{
if( buffer[i] == 0 )
buffer[i] = ',';
}
Now, instead of being separated by null characters, the attributes are separated by commas.
Actually, since I'm going to send it to Python I don't have to process it C-style after all. Just use the Py_BuildValue passing it the format character s#, which knows what do with it. You'll also need the size.
return Py_BuildValue("s#", buffer, size);
You can process it into a list on Python's end using split('\x00'). I found this after trial and error, but I'm glad to have learned something about C.