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Why do I get a segmentation fault when writing to a "char *s" initialized with a string literal, but not "char s[]"?
(19 answers)
Closed 1 year ago.
I'm trying to 'deep copy' a string so that I can perform operations on one copy, while retaining the original copy. This is the base example that I've got and for some reason the strncpy call causes a segfault. Please help
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
int main() {
char* stringA = "someVeryinTeresTingString";
char* stringB = malloc(sizeof(char) * strlen(stringA));
printf("A: %s, B: %s\n", stringA, stringB);
for (int i = 0; i < strlen(stringA); i++) {
stringB[i] = tolower(stringA[i]);
}
printf("A: %s, B: %s\n", stringA, stringB);
strncpy(stringA, stringB, strlen(stringA) - 1);
printf("A: %s, B: %s\n", stringA, stringB);
}
Easiest fix is to make a local copy of that string literal:
char stringA[] = "someVeryinTeresTingString";
Everything else works just the same.
Note that in the original code you have a pointer to immutable memory, while in this version you have a local (stack) array that is initialized with a copy of that string.
Another thing to note is if you're copying and manipulating C strings, do things like this:
char* stringB = strdup(stringA);
for (int i = 0; i < strlen(stringB); ++i) {
stringB[i] = tolower(stringB[i]);
}
Or even more efficiently by avoiding all these expensive strlen() calls:
char* stringB = strdup(stringA);
for (char* p = stringB; *p; ++p) {
*p = tolower(*p);
}
This line:
char* stringB = malloc(sizeof(char) * strlen(stringA));
shuld be like this:
char* stringB = malloc(sizeof(char) * (strlen(stringA) + 1));
then you will able to copy the \0 in the end of stringA
also, you want to copy to literal string - that is segmentation fault
char *strncpy(char *dest, const char *src, size_t n)
I'll try to comment and correct in your own code the mistakes I've seen:
(I will not correct things that can be eliminated or better done in another way, but are correct or not harmful, so you'll see only what must be corrected because of programming errors, and not questions about style or programming uses)
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
int main() {
char* stringA = "someVeryinTeresTingString";
/* you need to consider the space for the final null character in the malloc() call */
char* stringB = malloc(sizeof(char) * (strlen(stringA) + 1));
/* you don't need to use sizeof(char) as it is always equal to one.
* Multiplying by one is not necessary, but you'll probably know.
* char is warranteed by C standard that its sizeof is one. */
/* you need to copy the string *before* printing, or you will print an
* uninitialized string. Or at least initialize stringB to zeros, so you can
* use it with printf like functions (I do initialize the first char position to
* zero to make it appear as a length zero "" string)
* You will incurr in undefined behaviour if you don't do this. */
stringB[0] = '\0';
printf("A: %s, B: %s\n", stringA, stringB);
/* you need to copy the strings, so you can do it better if you test when
* stringA[i] == '\0', so you don't calculate the length of a string that is
* not going to change at every loop iteration. I will not change your
* code, because this is not an error. But strlen() searches from the
* beginning of the string for the '\0' char, character by character,
* and this test is done at every loop iteration. With the expression
* stringA[i] == 0 you do only a test per loop iteration to see if
* the char at position i in stringA is the null character. */
int i;
for (i = 0; i < strlen(stringA); i++) {
stringB[i] = tolower(stringA[i]);
}
/* you have not copied the final '\0', so I do it now. I need to move the
* declaration of i outside of the loop to be able to use it's value. */
stringB[i] = 0; /* you can use 0 or '\0' interchangeably */
printf("A: %s, B: %s\n", stringA, stringB);
/* nope. you need to copy the strings with a normal strcpy() as you know that
* both are the same length (better, you know that the space in stringB
* is the same as the length of stringA plus one). If you do this, you will not copy the last '\0' char, so wee need to append it.
* well, I don't know if that is what you want, so I don't actually touch anything here. */
strncpy(stringA, stringB, strlen(stringA) - 1);
/* stringB should be one char shorter than stringA */
printf("A: %s, B: %s\n", stringA, stringB);
}
by the way, you have been recommended to use strdup(3). This is a good idea, you don't need to be thinking on final nulls in this case, because strdup() takes care of it. Just remember that strdup(3) is not included in many C standard revisions, so you can get in trouble if you
move your program to a place lacking it (that should be very strange, anyway)
Related
I need help with char array. I want to create a n-lenght array and initialize its values, but after malloc() function the array is longer then n*sizeof(char), and the content of array isnt only chars which I assign... In array is few random chars and I dont know how to solve that... I need that part of code for one project for exam in school, and I have to finish by Sunday... Please help :P
#include<stdlib.h>
#include<stdio.h>
int main(){
char *text;
int n = 10;
int i;
if((text = (char*) malloc((n)*sizeof(char))) == NULL){
fprintf(stderr, "allocation error");
}
for(i = 0; i < n; i++){
//text[i] = 'A';
strcat(text,"A");
}
int test = strlen(text);
printf("\n%d\n", test);
puts(text);
free(text);
return 0;
}
Well before using strcat make
text[0]=0;
strcat expects null terminated char array for the first argument also.
From standard 7.24.3.1
#include <string.h>
char *strcat(char * restrict s1,
const char * restrict s2);
The strcat function appends a copy of the string pointed to by s2
(including the terminating null character) to the end of the string
pointed to by s1. The initial character of s2 overwrites the null
character at the end of s1.
How do you think strcat will know where the first string ends if you don't
put a \0 in s1.
Also don't forget to allocate an extra byte for the \0 character. Otherwise you are writing past what you have allocated for. This is again undefined behavior.
And earlier you had undefined behavior.
Note:
You should check the return value of malloc to know whether the malloc invocation was successful or not.
Casting the return value of malloc is not needed. Conversion from void* to relevant pointer is done implicitly in this case.
strlen returns size_t not int. printf("%zu",strlen(text))
To start with, you're way of using malloc in
text = (char*) malloc((n)*sizeof(char)
is not ideal. You can change that to
text = malloc(n * sizeof *text); // Don't cast and using *text is straighforward and easy.
So the statement could be
if(NULL == (text = (char*) malloc((n)*sizeof(char))){
fprintf(stderr, "allocation error");
}
But the actual problem lies in
for(i = 0; i < n; i++){
//text[i] = 'A';
strcat(text,"A");
}
The strcat documentation says
dest − This is pointer to the destination array, which should contain
a C string, and should be large enough to contain the concatenated
resulting string.
Just to point out that the above method is flawed, you just need to consider that the C string "A" actually contains two characters in it, A and the terminating \0(the null character). In this case, when i is n-2, you have out of bounds access or buffer overrun1. If you wanted to fill the entire text array with A, you could have done
for(i = 0; i < n; i++){
// Note for n length, you can store n-1 chars plus terminating null
text[i]=(n-2)==i?'A':'\0'; // n-2 because, the count starts from zero
}
//Then print the null terminated string
printf("Filled string : %s\n",text); // You're all good :-)
Note: Use a tool like valgrind to find memory leaks & out of bound memory accesses.
I was messing around with all of the string functions today and while most worked as expected, especially because I stopped trying to modify literals (sigh), there is one warning and oddity I can't seem to fix.
#include <stdio.h>
#include <string.h>
int main() {
char array[] = "Longword";
char *string = "Short";
strcpy(array, string); // Short
strcat(array, " "); // Short (with whitespace)
strcat(array, string); // Short Short
strtok(array, " "); // Short
if (strcmp(array, string) == 0)
{
printf("They are the same!\n");
}
char *substring = "or";
if (strstr(array, substring) != NULL)
{
printf("There's a needle in there somewhere!\n");
char *needle = strstr(array, substring);
int len = strlen(needle);
needle[len] = "\0"; // <------------------------------------------------
printf("Found it! There ya go: %s",needle);
}
printf("%s\n", array);
return 0;
}
Feel free to ignore the first few operations - I left them in because they modified array in a way, that made the strstr function useful to begin with.
The point in question is the second if statement, line 32 if you were to copy it in an editor.
(EDIT: Added arrow to the line. Sorry about that!)
This line is wrong:
needle[len] = "\0";
Doublequotes make a string literal, whose type is char *. But needle[len] is a char. To make a char literal you use singlequotes:
needle[len] = '\0';
See Single quotes vs. double quotes in C or C++
Your second strcat call overruns the end of array, corrupting whatever happens to be after it in memory. Once that happens, the later code might do just about anything, which is why writing past the end of an array is undefined behavior
So I'm working through "Sams Teach Yourself C Programming in One Hour a Day, Seventh Edition" Lesson 10 Exercise 7 which asks to "Write a function that accepts two strings. Use the malloc() function to allocate enough memory to hold the two strings after they have been concatenated (linked). Return a pointer to this new string."
I am sure there are much more elegant ways to go about this than what I have attempted below. I am mostly interested in why my solution doesn't work. I have only been learning C for a few months and have no significant programming background. Please let me know why this crashes on compilation. I am using Code Blocks on Win 7 with GNU GCC Compiler if that makes a difference. Thank you :)
#include <stdio.h>
#include <stdlib.h>
char * concatenated(char array1[], char array2[]);
int ctrtotal;
int main(void)
{
char *comboString;
char *array1 = "You\'re the man ";
char *array2 = "Now Dog!";
comboString = (char *)malloc(ctrtotal * sizeof(char));
concatenated(array1, array2);
if (comboString == NULL)
{
puts("Memory error");
exit(1);
}
puts(comboString);
free(comboString);
return 0;
}
char * concatenated(char array1[], char array2[])
{
char *array3;
int ctr;
int ctr2;
for (ctr = 0; array1[ctr] != '\0'; ctr++)
array3[ctr] = array1[ctr];
ctr2 = ctr;
for (ctr = 0; array2[ctr] != '\0'; ctr++)
{
array3[ctr2 + ctr] = array2[ctr];
}
array3[ctr2 + ctr + 1] = '\0';
ctrtotal = (ctr2 + ctr + 2);
return array3;
}
Thank you for the help. After reviewing everyone's feedback on my errors I revised the code to the following:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char * concatenated(char array1[], char array2[]);
int main(void)
{
char *array1 = "Testing Testing One Two ";
char *array2 = "Three. Finally, not crashing the mem o ry.";
char *comboString = malloc( (strlen(array1)+strlen(array2) + 1)*sizeof(char));
comboString = concatenated(array1, array2);
if (comboString == NULL)
{
puts("Memory error");
exit(1);
}
puts(comboString);
free(comboString);
return 0;
}
char * concatenated(char array1[], char array2[])
{
char *array3;
array3 = malloc( (strlen(array1)+strlen(array2) + 1)*sizeof(char) );
strcat(array3, array1);
strcat(array3, array2);
return array3;
}
If anyone sees any redundancies/unnecessary remaining code the could/should be deleted, please let me know. I recognize the benefit of being as concise as possible.
Your code has a bunch of issues:
int ctrtotal is never initialized, so you are mallocing 0 bytes
concatenated() is copying characters to an uninitialized array3. This pointer should point to a mallocd buffer.
If concatenated is allocating the memory, then main doesn't need to. Instead it should use the result of concatenated.
I don't want to give you the full code, and let you to miss out on this learning opportunity. So concatenated should look like this, in psuedo-code:
count = length_of(string1) + length_of(string2) + 1
buffer = malloc(count)
copy string1 to buffer
copy string2 to buffer, after string1
set the last byte of buffer to '\0' (NUL)
return buffer
In C, strings are represented as a NUL-terminated array of characters. That's why we allocate one additional byte, and terminate it with \0.
As a side-note, when dealing with strings, it is far easier to work with pointers, instead of treating them as arrays and accessing them via indices.
There's a lot of code here that just doesn't make any sense. I suggest that you first write this program on paper. Then, "execute" the program in your head, stepping through every line. If you get to something you don't understand, then you need to either fix your understanding, or your incorrect code. Don't try to write code that looks like some other bit of code.
There's also a library function called strcat which will make this task even easier. See if you can figure out how to use it here.
Spoiler --> #include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *concatenate2(const char* s1, const char* s2);
int main(void)
{
char *comboString;
char *array1 = "You're the man ";
char *array2 = "Now Dog!";
comboString = concatenate2(array1, array2);
if (comboString == NULL)
{
puts("Memory error");
exit(1);
}
puts(comboString);
free(comboString);
return 0;
}
char *concatenate2(const char* s1, const char* s2)
{
char *result;
result = malloc(strlen(s1) + strlen(s2) + 1);
*result = '\0';
strcat(result, s1);
strcat(result, s2);
return result;
}
You forgot to allocate memory for third, concatenated, array of chars (in function)
You should do something like this:
char *array3;
array3 = (char *)malloc( (strlen(array1)+strlen(array2) + 1)*sizeof(char) ); // +1 for '\0' character.
and then write chars from first and second array into third.
Perhaps a stroll through the question code is best.
#include <stdio.h>
#include <stdlib.h>
char * concatenated(char array1[], char array2[]);
int ctrtotal;
Notice that the above line declares ctrtotal to be an integer, but does not specify the value of the integer.
int main(void)
{
char *comboString;
char *array1 = "You\'re the man ";
char *array2 = "Now Dog!";
comboString = (char *)malloc(ctrtotal * sizeof(char));
Notice that the above line allocates memory and sets 'comboString' to point at that memory. However, how much memory is being allocated?
(ctrtotal[???] * sizeof(char)[1])
What is the value of (??? * 1) ? This is a problem.
concatenated(array1, array2);
The intent of the line above is that array1["You\'re the man "] and array2["Now Dog!"] will be joined to form a new string["You\'re the man Now Dog!"], which will be placed in allocated memory and returned to the caller.
Unfortunately, the returned memory containing the string is not captured here. For example, perhaps the above line should be:
comboString = concatenated(array1, array2);
While this make sense, for this line, it begs a question of the purpose of the lines:
comboString = (char *)malloc(ctrtotal * sizeof(char));
as well as the global variable:
int ctrtotal;
and the later reference:
ctrtotal = (ctr2 + ctr + 2);
Perhaps all of these 3 lines should be deleted?
if (comboString == NULL)
{
puts("Memory error");
exit(1);
}
puts(comboString);
free(comboString);
return 0;
}
char * concatenated(char array1[], char array2[])
{
char *array3;
Notice that '*array3' is now a defined pointer, but it is not pointing anywhere specific.
int ctr;
int ctr2;
The purpose of 'concatenated()' is to join array1 and array1 into allocated array3. Unfortunately, no memory is allocated to array3.
Below, the memory where array3 is pointing will be modified. Since array3 is not pointing anywhere specific, this is not safe.
Prior to modifying memory where array 3 is pointing, it is important to point array3 at memory where it is safe to modify bytes. I suggest that the following code be inserted here:
array3 = malloc(strlen(array1) + strlen(array2) + 1);
Now, array3 points to allocated memory, large enough to hold both strings plus the string termination character '\0'.
for (ctr = 0; array1[ctr] != '\0'; ctr++)
array3[ctr] = array1[ctr];
ctr2 = ctr;
for (ctr = 0; array2[ctr] != '\0'; ctr++)
{
array3[ctr2 + ctr] = array2[ctr];
}
array3[ctr2 + ctr + 1] = '\0';
ctrtotal = (ctr2 + ctr + 2);
return array3;
}
I am responding to your revised code. There are a few bugs in it.
...
char *array2 = "Three. Finally, not crashing the mem o ry.";
char *comboString = malloc( (strlen(array1)+strlen(array2) + 1)*sizeof(char));
comboString = concatenated(array1, array2);
...
The malloc is unnecessary here and actually a bug in your code. You are allocating a block of memory, but you then replace the value of the pointer comboString with the pointer from the call to concatenated. You lose the pointer to the block of memory allocated in main and thus never are able to free it. Although this will not be a problem in the code you have right now since main returns soon after, it could cause a memory leak in an application that ran for a longer time.
strcat(array3, array1);
This is also a bug. strcat is going to walk through array3 to find '\0' and then once it is found copy in array1 from that index on, replacing the '\0'. This works fine here since the memory block that was allocated for array3 is going to be zeroed out** as no block has yet been freed by your program. However, in a longer running program you can end up with a block that does not start with a '\0'. You might end up corrupting your heap, getting a segfault, etc.
To fix this, you should use strcpy instead, array3[0] = '\0', or *array3 = '\0'
** When the operating system starts your program it will initialize the memory segment it reserves for it with zeroes (this actually isn't a necessity but will be true on almost any operating system). As your program allocates and frees memory, you will eventually wind up with values that are not zero. Note that the same bug can occur with uninitialized local variables such as:
int i;
for (; i < 10; i++);
This loop will run 10 times whenever the space on the runtime stack where i is stored is already 0.
Overall, the takeaway is to be very careful with arrays and dynamic memory allocation in C. C offers you none of the protections that modern languages do. You are responsible for making sure you stay within the bounds of your array, initialize your variables, and properly allocate and free your memory. Neglecting these things will lead to obscure bugs that will take you hours to find, and most of the times these bugs will not appear right away.
I am having trouble concatenating strings in C, without strcat library function. Here is my code
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
char *a1=(char*)malloc(100);
strcpy(a1,"Vivek");
char *b1=(char*)malloc(100);
strcpy(b1,"Ratnavel");
int i;
int len=strlen(a1);
for(i=0;i<strlen(b1);i++)
{
a1[i+len]=b1[i];
}
a1[i+len]='\0';
printf("\n\n A: %s",a1);
return 0;
}
I made corrections to the code. This is working. Now can I do it without strcpy?
Old answer below
You can initialize a string with strcpy, like in your code, or directly when declaring the char array.
char a1[100] = "Vivek";
Other than that, you can do it char-by-char
a1[0] = 'V';
a1[1] = 'i';
// ...
a1[4] = 'k';
a1[5] = '\0';
Or you can write a few lines of code that replace strcpy and make them a function or use directly in your main function.
Old answer
You have
0 1 2 3 4 5 6 7 8 9 ...
a1 [V|i|v|e|k|0|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_]
b1 [R|a|t|n|a|v|e|l|0|_|_|_|_|_|_|_|_|_|_|_|_|_]
and you want
0 1 2 3 4 5 6 7 8 9 ...
a1 [V|i|v|e|k|R|a|t|n|a|v|e|l|0|_|_|_|_|_|_|_|_]
so ...
a1[5] = 'R';
a1[6] = 'a';
// ...
a1[12] = 'l';
a1[13] = '\0';
but with loops and stuff, right? :D
Try this (remember to add missing bits)
for (aindex = 5; aindex < 14; aindex++) {
a1[aindex] = b1[aindex - 5];
}
Now think about the 5 and 14 in the loop above.
What can you replace them with? When you answer this, you have solved the programming problem you have :)
char a1[] = "Vivek";
Will create a char array a1 of size 6. You are trying to stuff it with more characters than it can hold.
If you want to be able to accommodate concatenation "Vivek" and "Ratnavel" you need to have a char array of size atleast 14 (5 + 8 + 1).
In your modified program you are doing:
char *a1=(char*)malloc(100); // 1
a1 = "Vivek"; // 2
1: Will allocate a memory chunk of size 100 bytes, makes a1 point to it.
2: Will make a1 point to the string literal "Vivek". This string literal cannot be modified.
To fix this use strcpy to copy the string into the allocated memory:
char *a1=(char*)malloc(100);
strcpy(a1,"Vivek");
Also the for loop condition i<strlen(b1)-1 will not copy last character from the string, change it to i<strlen(b1)
And
a1[i]='\0';
should be
a1[i + len]='\0';
as the new length of a1 is i+len and you need to have the NUL character at that index.
And don't forget to free your dynamically allocated memory once you are done using it.
You cannot safely write into those arrays, since you have not made sure that enough space is available. If you use malloc() to allocate space, you can't then overwrite the pointer by assigning to string literal. You need to use strcpy() to copy a string into the newly allocated buffers, in that case.
Also, the length of a string in C is computed by the strlen() function, not length() that you're using.
When concatenating, you need to terminate at the proper location, which your code doesn't seem to be doing.
Here's how I would re-implement strcat(), if needed for some reason:
char * my_strcat(char *out, const char *in)
{
char *anchor = out;
size_t olen;
if(out == NULL || in == NULL)
return NULL;
olen = strlen(out);
out += olen;
while(*out++ = *in++)
;
return anchor;
}
Note that this is just as bad as strcat() when it comes to buffer overruns, since it doesn't support limiting the space used in the output, it just assumes that there is enough space available.
Problems:
length isn't a function. strlen is, but you probably shouldn't call it in a loop - b1's length won't change on us, will it? Also, it returns a size_t, which may be the same size as int on your platform but will be unsigned. This can (but usually won't) cause errors, but you should do it right anyway.
a1 only has enough space for the first string, because the compiler doesn't know to allocate extra stack space for the rest of the string since. If you provide an explicit size, like [100], that should be enough for your purposes. If you need robust code that doesn't make assumptions about what is "enough", you should look into malloc and friends, though that may be a lesson for another day.
Your loop stops too early. i < b1_len (assuming you have a variable, b1_len, that was set to the length of b1 before the loop began) would be sufficient - strlen doesn't count the '\0' at the end.
But speaking of counting the '\0' at the end, a slightly more efficient implementation could use sizeof a1 - 1 instead of strlen(a1) in this case, since a1 (and b1) are declared as arrays, not pointers. It's your choice, but remember that sizeof won't work for pointers, so don't get them mixed up.
EDIT: New problems:
char *p = malloc(/*some*/); p = /* something */ is a problem. = with pointers doesn't copy contents, it copies the value, so you're throwing away the old pointer value you got from malloc. To copy the contents of a string into a char * (or a char [] for that matter) you'd need to use strcpy, strncpy, or (my preference) memcpy. (Or just a loop, but that's rather silly. Then again, it may be good practice if you're writing your own strcat.)
Unless you're using C++, I wouldn't cast the return value of malloc, but that's a religious war and we don't need one of those.
If you have strdup, use it. If you don't, here is a working implementation:
char *strdup(const char *c)
{
size_t l = strlen(c);
char *d = malloc(l + 1);
if(d) memcpy(d, c, l + 1);
return d;
}
It is one of the most useful functions not in the C standard library.
You can do it using strcpy() too ;)
char *a = (char *) malloc(100);
char *b = (char *) malloc(100);
strcpy(a, "abc"); // initializes a
strcpy(b, "def"); // and b
strcpy((a + strlen(a)), b); // copy b at end of a
printf("%s\n",a); // will produce: "abcdef"
i think this is an easy one.
#include<stdio.h>
int xstrlen(char *);
void xstrcat(char *,char *,int);
void main()
{
char source[]="Sarker";
char target[30]="Maruf";
int j=xstrlen(target);
xstrcat(target,source,j);
printf("Source String: %s\nTarget String: %s",source,target);
}
int xstrlen(char *s)
{
int len=0;
while(*s!='\0')
{
len++;
s++;
}
return len;
}
void xstrcat(char *t,char *s,int j)
{
while(*t!='\0')
{
*t=*t;
t++;
}
while(*s!='\0')
{
*t=*s;
s++;
t++;
}
}
It is better to factor out your strcat logic to a separate function. If you make use of pointer arithmetic, you don't need the strlen function:
#include <stdio.h>
#include <stdlib.h>
#include <string.h> /* To completely get rid of this,
implement your our strcpy as well */
static void
my_strcat (char* dest, char* src)
{
while (*dest) ++dest;
while (*src) *(dest++) = *(src++);
*dest = 0;
}
int
main()
{
char* a1 = malloc(100);
char* b1 = malloc(100);
strcpy (a1, "Vivek");
strcpy (b1, " Ratnavel");
my_strcat (a1, b1);
printf ("%s\n", a1); /* => Vivek Ratnavel */
free (a1);
free (b1);
return 0;
}
I trying to do some very basic string processing in C (e.g. given a filename, chop off the file extension, manipulate filename and then add back on the extension)- I'm rather rusty on C and am getting segmentation faults.
char* fname;
char* fname_base;
char* outdir;
char* new_fname;
.....
fname = argv[1];
outdir = argv[2];
fname_len = strlen(fname);
strncpy(fname_base, fname, (fname_len-4)); // weird characters at the end of the truncation?
strcpy(new_fname, outdir); // getting a segmentation on this I think
strcat(new_fname, "/");
strcat(new_fname, fname_base);
strcat(new_fname, "_test");
strcat(new_fname, ".jpg");
printf("string=%s",new_fname);
Any suggestions or pointers welcome.
Many thanks and apologies for such a basic question
You need to allocate memory for new_fname and fname_base. Here's is how you would do it for new_fname:
new_fname = (char*)malloc((strlen(outdir)+1)*sizeof(char));
In strlen(outdir)+1, the +1 part is for allocating memory for the NULL CHARACTER '\0' terminator.
In addition to what other's are indicating, I would be careful with
strncpy(fname_base, fname, (fname_len-4));
You are assuming you want to chop off the last 4 characters (.???). If there is no file extension or it is not 3 characters, this will not do what you want. The following should give you an idea of what might be needed (I assume that the last '.' indicates the file extension). Note that my 'C' is very rusty (warning!)
char *s;
s = (char *) strrchr (fname, '.');
if (s == 0)
{
strcpy (fname_base, fname);
}
else
{
strncpy (fname_base, fname, strlen(fname)-strlen(s));
fname_base[strlen(fname)-strlen(s)] = 0;
}
You have to malloc fname_base and new_fname, I believe.
ie:
fname_base = (char *)(malloc(sizeof(char)*(fname_len+1)));
fname_base[fname_len] = 0; //to stick in the null termination
and similarly for new_fname and outdir
You're using uninitialized pointers as targets for strcpy-like functions: fname_base and new_fname: you need to allocate memory areas to work on, or declare them as char array e.g.
char fname_base[FILENAME_MAX];
char new_fname[FILENAME_MAX];
you could combine the malloc that has been suggested, with the string manipulations in one statement
if ( asprintf(&new_fname,"%s/%s_text.jpg",outdir,fname_base) >= 0 )
// success, else failed
then at some point, free(new_fname) to release the memory.
(note this is a GNU extension which is also available in *BSD)
Cleaner code:
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
const char *extra = "_test.jpg";
int main(int argc, char** argv)
{
char *fname = strdup(argv[1]); /* duplicate, we need to truncate the dot */
char *outdir = argv[1];
char *dotpos;
/* ... */
int new_size = strlen(fname)+strlen(extra);
char *new_fname = malloc(new_size);
dotpos = strchr(fname, '.');
if(dotpos)
*dotpos = '\0'; /* truncate at the dot */
new_fname = malloc(new_size);
snprintf(new_fname, new_size, "%s%s", fname, extra);
printf("%s\n", new_fname);
return 0;
}
In the following code I do not call malloc.
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
/* Change this to '\\' if you are doing this on MS-windows or something like it. */
#define DIR_SYM '/'
#define EXT_SYM '.'
#define NEW_EXT "jpg"
int main(int argc, char * argv[] ) {
char * fname;
char * outdir;
if (argc < 3) {
fprintf(stderr, "I want more command line arguments\n");
return 1;
}
fname = argv[1];
outdir = argv[2];
char * fname_base_begin = strrchr(fname, DIR_SYM); /* last occurrence of DIR_SYM */
if (!fname_base_begin) {
fname_base_begin = fname; // No directory symbol means that there's nothing
// to chop off of the front.
}
char * fname_base_end = strrchr(fname_base_begin, EXT_SYM);
/* NOTE: No need to search for EXT_SYM in part of the fname that we have cut off
* the front and then have to deal with finding the last EXT_SYM before the last
* DIR_SYM */
if (!fname_base_end) {
fprintf(stderr, "I don't know what you want to do when there is no extension\n");
return 1;
}
*fname_base_end = '\0'; /* Makes this an end of string instead of EXT_SYM */
/* NOTE: In this code I actually changed the string passed in with the previous
* line. This is often not what you want to do, but in this case it should be ok.
*/
// This line should get you the results I think you were trying for in your example
printf("string=%s%c%s_test%c%s\n", outdir, DIR_SYM, fname_base_begin, EXT_SYM, NEW_EXT);
// This line should just append _test before the extension, but leave the extension
// as it was before.
printf("string=%s%c%s_test%c%s\n", outdir, DIR_SYM, fname_base_begin, EXT_SYM, fname_base_end+1);
return 0;
}
I was able to get away with not allocating memory to build the string in because I let printf actually worry about building it, and took advantage of knowing that the original fname string would not be needed in the future.
I could have allocated the space for the string by calculating how long it would need to be based on the parts and then used sprintf to form the string for me.
Also, if you don't want to alter the contents of the fname string you could also have used:
printf("string=%s%c%*s_test%c%s\n", outdir, DIR_SYM, (unsigned)fname_base_begin -(unsigned)fname_base_end, fname_base_begin, EXT_SYM, fname_base_end+1);
To make printf only use part of the string.
The basic of any C string manipulation is that you must write into (and read from unless... ...) memory you "own". Declaring something is a pointer (type *x) reserves space for the pointer, not for the pointee that of course can't be known by magic, and so you have to malloc (or similar) or to provide a local buffer with things like char buf[size].
And you should be always aware of buffer overflow.
As suggested, the usage of sprintf (with a correctly allocated destination buffer) or alike could be a good idea. Anyway if you want to keep your current strcat approach, I remember you that to concatenate strings, strcat have always to "walk" thourgh the current string from its beginning, so that, if you don't need (ops!) buffer overflow checks of any kind, appending chars "by hand" is a bit faster: basically when you finished appending a string, you know where the new end is, and in the next strcat, you can start from there.
But strcat doesn't allow to know the address of the last char appended, and using strlen would nullify the effort. So a possible solution could be
size_t l = strlen(new_fname);
new_fname[l++] = '/';
for(i = 0; fname_base[i] != 0; i++, l++) new_fname[l] = fname_base[i];
for(i = 0; testjpgstring[i] != 0; i++, l++) new_fname[l] = testjpgstring[i];
new_fname[l] = 0; // terminate the string...
and you can continue using l... (testjpgstring = "_test.jpg")
However if your program is full of string manipulations, I suggest using a library for strings (for lazyness I often use glib)