I have a global definition as following:
#define globalstring "example1"
typedef struct
{
char key[100];
char trail[10][100];
bson_value_t value;
} ObjectInfo;
typedef struct
{
ObjectInfo CurrentOrderInfoSet[5];
} DataPackage;
DataPackage GlobalDataPackage[10];
And I would like to use the strcpy() function in some of my functions as following:
strcpy(GlobalDataPackage[2].CurrentOrderInfoSet[0].key, "example2");
char string[100] = "example3";
strcpy(GlobalDataPackage[2].CurrentOrderInfoSet[0].key, string);
strcpy(GlobalDataPackage[2].CurrentOrderInfoSet[0].key, globalstring);
First question: Are the global defined strings all initiated with 100 times '\0'?
Second qestion: I am a bit confused as to how exactly strcpy() works. Does it only overwrite the characters necessary to place the source string into the destination string plus a \0 at the end and leave the rest as it is or does it fully delete any content of the destination string prior to that?
Third question: All my strings are fixed length of 100. If I use the 3 examples of strcpy() above, with my strings not exceeding 99 characters, does strcpy() properly overwrite the destination string and NULL terminate it? Meaning do I run into problems when using functions like strlen(), printf() later?
Fourth question: What happens when I strcpy() empty strings?
I plan to overwrite these strings in loops various times and would like to know if it would be safer to use memset() to fully "empty" the strings prior to strcpy() on every iteration.
Thx.
Are the global defined strings all initiated with 100 times '\0'?
Yes. Global char arrays will be initilizated to all zeros.
I am a bit confused as to how exactly strcpy() works. Does it only overwrite the characters necessary to place the source string into the destination string plus a \0 at the end and leave the rest as it
Exactly. It copies the characters up until and including '\0' and does not care about the rest.
If I use ... my strings not exceeding 99 characters, does strcpy() properly overwrite the destination string and NULL terminate it?
Yes, but NULL is a pointer, it's terminated with zero byte, sometimes called NUL. You might want to see What is the difference between NUL and NULL? .
Meaning do I run into problems when using functions like strlen(), printf() later?
Not if your string lengths are less than or equal to 99.
What happens when I strcpy() empty strings?
It just copies one zero byte.
would like to know if it would be safer to use memset() to fully "empty" the strings prior to strcpy() on every iteration.
Safety is a broad concept. As far as safety as in if the program will execute properly, there is no point in caring about anything after zero byte, so just strcpy it.
But you should check if your strings are less than 99 characters and handle what to do it they are longer. You might be interested in strnlen, but the interface is confusing - I recommend to use memcpy + explicitly manually set zero byte.
My code is crashing because of a lack of the char '\0' at the end of some strings.
It's pretty clear to me why we have to use this termination char. My question is,
is there a problem adding a potential 2nd null character to a character array - to solve string problems?
I think it's cheaper just add a '\0' to every string than verify if it needs and then add it, but I don't know if it's a good thing to do.
is there a problem to have this char ('\0') twice at the end of a string?
This question lacks clarity as "string" means different things to people.
Let us use the C specification definition as this is a C post.
A string is a contiguous sequence of characters terminated by and including the first null character. C11 ยง7.1.1 1
So a string, cannot have 2 null characters as the string ends upon reaching its first one. #Michael Walz
Instead, re-parse to "is there a problem adding a potential 2nd null character to a character array - to solve string problems?"
A problem with attempting to add a null character to a string is confusion. The str...() functions work with C strings as defined above.
// If str1 was not a string, strcpy(str1, anything) would be undefined behavior.
strcpy(str1, "\0"); // no change to str1
char str2[] = "abc";
str2[strlen(str2)] = '\0'; // OK but only, re-assigns the \0 to a \0
// attempt to add another \0
str2[strlen(str2)+1] = '\0'; // Bad: assigning outside `str2[]` as the array is too small
char str3[10] = "abc";
str3[strlen(str3)+1] = '\0'; // OK, in this case
puts(str3); // Adding that \0 served no purpose
As many have commented, adding a spare '\0' is not directly attending the code's fundamental problem. #Haris #Malcolm McLean
That unposted code is the real issue that need solving #Yunnosch, and not by attempting to append a second '\0'.
I think it's cheaper just add a '\0' to every string than verify if it needs and then add it, but I don't know if it's a good thing to do.
Where would you add it? Let's assume we've done something like this:
char *p = malloc(32);
Now, if we know the allocated length, we could put a '\0' as the last character of the allocated area, as in p[31] = '\0'. But we don't how long the contents of the string are supposed to be. If there's supposed to be just foobar, then there'd still be 25 bytes of garbage, which might cause other issues if processed or printed.
Let alone the fact that if all you have is the pointer to the string, it's hard to know the length of the allocated area.
Probably better to fix the places where you build the strings to do it correctly.
Having '\0' is not a problem, unless you have not gone out of bounds of that char array.
You do have to understand that, having '\0' twice would mean, any string operation would not even know that there is a second '\0'. They will just read till the first '\0', and be with it. For them, the first '\0' is the Null terminating character and there should not be anything after that.
I'm new to C++ and I have to use array. The problem is I get error "array bounds overflow" in this line:
char arr[2] = "12";
But when I changed it to:
char arr[3] = "12";
it works fine but why?
Update:
And this works :(
char arr[2] = {'1','2'};
I'm really confused about the difference between declarations, how they are stored in the memory.
In the C family of languages the memory spaces which represent strings ( char arrays ) are terminated by the null character \0
Thus the memory to store the string must be at least one character larger than the expected size when you write it out with " "
Your new example, isn't creating a string, but rather an array of characters. Because you have switched notations form " " to { } the system is no longer creating a null terminated string but is rather creating an array as that is what you have asked for.
The crux of it is that Strings are special and have \0 tacked onto their end by the system automatically and therefore need additional space.
The char array has a null terminating character "\0" at the end of every string. You always need to reserve an additional space in your array for this character.
That's because literal strings in C and C++ have an implied '\0' appended to them. It's called a zero-terminated string, it helps when trying to keep track of the length of the string, instead of storing it explicitly somewhere in memory.
Because string constants must store a NUL at the end of string, 2 chars of storage is not enough, hence the overflow. You need to store '1', '2', and NUL which is 3 chars.
I have a little problem here with memcpy()
When I write this
char ipA[15], ipB[15];
size_t b = 15;
memcpy(ipA,line+15,b);
It copies b bytes from array line starting at 15th element (fine, this is what i want)
memcpy(ipB,line+31,b);
This copies b bytes from line starting at 31st element, but it also attaches to it the result for previous command i.e ipA.
Why? ipB size is 15, so it shouldnt have enough space to copy anything else. whats happening here?
result for ipA is 192.168.123.123
result for ipB becomes 205.123.123.122 192.168.123.123
Where am I wrong? I dont actually know alot about memory allocation in C.
It looks like you're not null-terminating the string in ipA. The compiler has put the two variables next to one another in memory, so string operations assume that the first null terminator is sometime after the second array (whenever the next 0 occurs in memory).
Try:
char ipA[16], ipB[16];
size_t b = 15;
memcpy(ipA,line+15,b);
ipA[15] = '\0';
memcpy(ipB,line+31,b);
ipB[15] = '\0';
printf("ipA: %s\nipB: %s\n", ipA, ipB)
This should confirm whether this is the problem. Obviously you could make the code a bit more elegant than my test code above. As an alternative to manually terminating, you could use printf("%.*s\n", b, ipA); or similar to force printf to print the correct number of characters.
Are you checking the content of the arrays by doing printf("%s", ipA) ? If so, you'll end up with the described effect since your array is interpreted as a C string which is not null terminated. Do this instead: printf("%.*s", sizeof(ipA), ipA)
Character strings in C require a terminating mark. It is the char value 0.
As your two character strings are contiguous in memory, if you don't terminate the first character string, then when reading it, you will continue until memory contains the end-of-string character.
I am new to C and I am very much confused with the C strings. Following are my questions.
Finding last character from a string
How can I find out the last character from a string? I came with something like,
char *str = "hello";
printf("%c", str[strlen(str) - 1]);
return 0;
Is this the way to go? I somehow think that, this is not the correct way because strlen has to iterate over the characters to get the length. So this operation will have a O(n) complexity.
Converting char to char*
I have a string and need to append a char to it. How can i do that? strcat accepts only char*. I tried the following,
char delimiter = ',';
char text[6];
strcpy(text, "hello");
strcat(text, delimiter);
Using strcat with variables that has local scope
Please consider the following code,
void foo(char *output)
{
char *delimiter = ',';
strcpy(output, "hello");
strcat(output, delimiter);
}
In the above code,delimiter is a local variable which gets destroyed after foo returned. Is it OK to append it to variable output?
How strcat handles null terminating character?
If I am concatenating two null terminated strings, will strcat append two null terminating characters to the resultant string?
Is there a good beginner level article which explains how strings work in C and how can I perform the usual string manipulations?
Any help would be great!
Last character: your approach is correct. If you will need to do this a lot on large strings, your data structure containing strings should store lengths with them. If not, it doesn't matter that it's O(n).
Appending a character: you have several bugs. For one thing, your buffer is too small to hold another character. As for how to call strcat, you can either put the character in a string (an array with 2 entries, the second being 0), or you can just manually use the length to write the character to the end.
Your worry about 2 nul terminators is unfounded. While it occupies memory contiguous with the string and is necessary, the nul byte at the end is NOT "part of the string" in the sense of length, etc. It's purely a marker of the end. strcat will overwrite the old nul and put a new one at the very end, after the concatenated string. Again, you need to make sure your buffer is large enough before you call strcat!
O(n) is the best you can do, because of the way C strings work.
char delimiter[] = ",";. This makes delimiter a character array holding a comma and a NUL Also, text needs to have length 7. hello is 5, then you have the comma, and a NUL.
If you define delimiter correctly, that's fine (as is, you're assigning a character to a pointer, which is wrong). The contents of output won't depend on delimiter later on.
It will overwrite the first NUL.
You're on the right track. I highly recommend you read K&R C 2nd Edition. It will help you with strings, pointers, and more. And don't forget man pages and documentation. They will answer questions like the one on strcat quite clearly. Two good sites are The Open Group and cplusplus.com.
A "C string" is in reality a simple array of chars, with str[0] containing the first character, str[1] the second and so on. After the last character, the array contains one more element, which holds a zero. This zero by convention signifies the end of the string. For example, those two lines are equivalent:
char str[] = "foo"; //str is 4 bytes
char str[] = {'f', 'o', 'o', 0};
And now for your questions:
Finding last character from a string
Your way is the right one. There is no faster way to know where the string ends than scanning through it to find the final zero.
Converting char to char*
As said before, a "string" is simply an array of chars, with a zero terminator added to the end. So if you want a string of one character, you declare an array of two chars - your character and the final zero, like this:
char str[2];
str[0] = ',';
str[1] = 0;
Or simply:
char str[2] = {',', 0};
Using strcat with variables that has local scope
strcat() simply copies the contents of the source array to the destination array, at the offset of the null character in the destination array. So it is irrelevant what happens to the source after the operation. But you DO need to worry if the destination array is big enough to hold the data - otherwise strcat() will overwrite whatever data sits in memory right after the array! The needed size is strlen(str1) + strlen(str2) + 1.
How strcat handles null terminating character?
The final zero is expected to terminate both input strings, and is appended to the output string.
Finding last character from a string
I propose a thought experiment: if it were generally possible to find the last character
of a string in better than O(n) time, then could you not also implement strlen
in better than O(n) time?
Converting char to char*
You temporarily can store the char in an array-of-char, and that will decay into
a pointer-to-char:
char delimiterBuf[2] = "";
delimiterBuf[0] = delimiter;
...
strcat(text, delimiterBuf);
If you're just using character literals, though, you can simply use string literals instead.
Using strcat with variables that has local scope
The variable itself isn't referenced outside the scope. When the function returns,
that local variable has already been evaluated and its contents have already been
copied.
How strcat handles null terminating character?
"Strings" in a C are NUL-terminated sequences of characters. Both inputs to
strcat must be NUL-terminated, and the result will be NUL-terminated. It
wouldn't be useful for strcat to write an extra NUL-byte to the result if it
doesn't need to.
(And if you're wondering what if the input strings have multiple trailing
NUL bytes already, I propose another thought experiment: how would strcat know
how many trailing NUL-bytes there are in a string?)
BTW, since you tagged this with "best-practices", I'll also recommend that you take care not to write past the end of your destination buffers. Typically this means avoiding strcat and strcpy (unless you've already checked that the input strings won't overflow the destination) and using safer versions (e.g. strncat. Note that strncpy has its own pitfalls, so that's a poor substitute. There also are safer versions that are non-standard, such as strlcpy/strlcat and strcpy_s/strcat_s.)
Similarly, functions like your foo function always should take an additional argument specifying what the size of the destination buffer is (and documentation should make it explicitly clear whether that size accounts for a NUL terminator or not).
How can I find out the last character
from a string?
Your technique with str[strlen(str) - 1] is fine. As pointed out, you should avoid repeated, unnecessary calls to strlen and store the results.
I somehow think that, this is not the
correct way because strlen has to
iterate over the characters to get the
length. So this operation will have a
O(n) complexity.
Repeated calls to strlen can be a bane of C programs. However, you should avoid premature optimization. If a profiler actually demonstrates a hotspot where strlen is expensive, then you can do something like this for your literal string case:
const char test[] = "foo";
sizeof test // 4
Of course if you create 'test' on the stack, it incurs a little overhead (incrementing/decrementing stack pointer), but no linear time operation involved.
Literal strings are generally not going to be so gigantic. For other cases like reading a large string from a file, you can store the length of the string in advance as but one example to avoid recomputing the length of the string. This can also be helpful as it'll tell you in advance how much memory to allocate for your character buffer.
I have a string and need to append a
char to it. How can i do that? strcat
accepts only char*.
If you have a char and cannot make a string out of it (char* c = "a"), then I believe you can use strncat (need verification on this):
char ch = 'a';
strncat(str, &ch, 1);
In the above code,delimiter is a local
variable which gets destroyed after
foo returned. Is it OK to append it to
variable output?
Yes: functions like strcat and strcpy make deep copies of the source string. They don't leave shallow pointers behind, so it's fine for the local data to be destroyed after these operations are performed.
If I am concatenating two null
terminated strings, will strcat
append two null terminating characters
to the resultant string?
No, strcat will basically overwrite the null terminator on the dest string and write past it, then append a new null terminator when it's finished.
How can I find out the last character from a string?
Your approach is almost correct. The only way to find the end of a C string is to iterate throught the characters, looking for the nul.
There is a bug in your answer though (in the general case). If strlen(str) is zero, you access the character before the start of the string.
I have a string and need to append a char to it. How can i do that?
Your approach is wrong. A C string is just an array of C characters with the last one being '\0'. So in theory, you can append a character like this:
char delimiter = ',';
char text[7];
strcpy(text, "hello");
int textSize = strlen(text);
text[textSize] = delimiter;
text[textSize + 1] = '\0';
However, if I leave it like that I'll get zillions of down votes because there are three places where I have a potential buffer overflow (if I didn't know that my initial string was "hello"). Before doing the copy, you need to put in a check that text is big enough to contain all the characters from the string plus one for the delimiter plus one for the terminating nul.
... delimiter is a local variable which gets destroyed after foo returned. Is it OK to append it to variable output?
Yes that's fine. strcat copies characters. But your code sample does no checks that output is big enough for all the stuff you are putting into it.
If I am concatenating two null terminated strings, will strcat append two null terminating characters to the resultant string?
No.
I somehow think that, this is not the correct way because strlen has to iterate over the characters to get the length. So this operation will have a O(n) complexity.
You are right read Joel Spolsky on why C-strings suck. There are few ways around it. The ways include either not using C strings (for example use Pascal strings and create your own library to handle them), or not use C (use say C++ which has a string class - which is slow for different reasons, but you could also write your own to handle Pascal strings more easily than in C for example)
Regarding adding a char to a C string; a C string is simply a char array with a nul terminator, so long as you preserve the terminator it is a string, there's no magic.
char* straddch( char* str, char ch )
{
char* end = &str[strlen(str)] ;
*end = ch ;
end++ ;
*end = 0 ;
return str ;
}
Just like strcat(), you have to know that the array that str is created in is long enough to accommodate the longer string, the compiler will not help you. It is both inelegant and unsafe.
If I am concatenating two null
terminated strings, will strcat append
two null terminating characters to the
resultant string?
No, just one, but what ever follows that may just happen to be nul, or whatever happened to be in memory. Consider the following equivalent:
char* my_strcat( char* s1, const char* s2 )
{
strcpy( &str[strlen(str)], s2 ) ;
}
the first character of s2 overwrites the terminator in s1.
In the above code,delimiter is a local
variable which gets destroyed after
foo returned. Is it OK to append it to
variable output?
In your example delimiter is not a string, and initialising a pointer with a char makes no sense. However if it were a string, the code would be fine, strcat() copies the data from the second string, so the lifetime of the second argument is irrelevant. Of course you could in your example use a char (not a char*) and the straddch() function suggested above.