strcat concat a char onto a string?

Question

Using GDB, I find I get a segmentation fault when I attempt this operation:

strcat(string,&currentChar);

Given that string is initialized as

char * string = "";

and currentChar is

char currentChar = 'B';

Why does this result in a segmentation fault?

If strcat can't be used for this, how else can I concat a char onto a string?

strcat requires null-terminated strings. Your single char has no null, so strcat will keep copying until it finds a null somewhere down the line, which could be on the 2nd byte by accident, or a few gigabytes down the road, all the while copying all that garbage into your string, stomping on the stack — Marc B, Jan 29 '11 at 02:52

gallo · Answer 1 · 2017-01-07T18:49:21.080

As responded by others, &currentChar is a pointer to char or char*, but a string in C is char[] or const char*.

One way to use strcat to concatenate a char to string is creating a minimum string and use it to transform a char into string.

Example:

Making a simple string, with only 1 character and the suffix '\0';

char cToStr[2];
cToStr[1] = '\0';

Applying to your question:

char * string = "";
char currentChar = 'B';

cToStr will assume the string "B":

cToStr[0] = currentChar;

And strcat will work!

strcat ( string, cToStr );

Elalfer · Accepted Answer · 2011-02-02T02:15:28.743

10

Because &currentChar is not a string, it doesn't finish with \0 character. You should define B as char *currentChar = 'B';. Also according to http://www.cplusplus.com/reference/clibrary/cstring/strcat string should have enough space to hold the result string (2 bytes in this case), but it is only 1 byte.

Or if you want to use char then you can do something like (depending of your code):

char string[256];
...

char currentChar = 'B';
size_t cur_len = strlen(string);
if(cur_len < 254) {
    string[cur_len] = currentChar;
    string[cur_len+1] = '\0';
}
else
    printf("Not enough space");

edited Feb 02 '11 at 02:15

answered Jan 29 '11 at 02:51

Elalfer

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Not the reason why the program is segfaulting. This is part of the error though – Foo Bah Jan 29 '11 at 03:00
@Foo `&currentChar` doesn't point on null terminates string and of course strcpy will cause a seg fault – Elalfer Jan 29 '11 at 03:03
I accepting this as the answer because it covers that A: I needed the null terminators, and B: I had to allocate more space for string. – Blackbinary Jan 29 '11 at 03:13
1

string[cur_len+1] should be set to 0, not '\n' – Ferruccio Feb 02 '11 at 01:24

score 3 · Answer 3 · answered Apr 05 '16 at 01:50

3

I think the simplest method (not efficient) would be sprintf

sprintf(str, "%s%c", str, chr);

answered Apr 05 '16 at 01:50

Nandakumar Edamana

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why is not efficient? – Gustavo Adolfo Mejía May 02 '18 at 16:54
1

@GustavoAdolfoMejía because it has to parse the format string. You can write your own function (say `strcatchar()`) to do the same job more efficiently. – Nandakumar Edamana May 04 '18 at 00:17
Also with some compiler you would get a warning: passing argument 1 to restrict-qualified parameter aliases with argument 3 – Alexandre May 31 '22 at 18:06

score 2 · Answer 4 · answered Jan 29 '11 at 02:52

strcat() takes two '\0'-terminated strings. When you pass the address of a character, the routine will look at the memory that follows the character, looking for the terminator.

Since you don't know what that memory even refers to, you should expect problems when your code accesses it.

In addition to that, your string argument does not have room to have any characters appended to it. Where is that memory written to? It will attempt to write past the end of the memory associated with this string.

score 0 · Answer 5 · edited May 22 '17 at 18:30

0

Both of the strings must be null-terminated. A single char isn't null terminated, so it's undefined when strcat will stop concatenating characters to the end. Also, string must contain at least enough space for both the original string and resultant string.

This works:

char string[10] = "";
char* currentChar = "B";
strcat(string, currentChar);

edited May 22 '17 at 18:30

Peter Mortensen

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answered Jan 29 '11 at 02:52

helloworld922

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This does not work because there is not enough space in string. To see my point, try it: http://ideone.com/0jjST – Foo Bah Jan 29 '11 at 02:54
There's two problems with the code. The first argument doesn't have enough space, the second one has no terminator. Why are you focused on only one issue? – Jonathan Wood Jan 29 '11 at 02:58
it's part of the reason, though you're right it's not the main one. fixed to include the main cause of the segfault – helloworld922 Jan 29 '11 at 02:58

score 0 · Answer 6 · answered Jan 28 '21 at 10:38

0

We know that currentChar = 'B'.

This can be done

strcat(string, "B\0");

If we know currentChar will be hardcoded as 'B', this would be a good approach.
It also removes the need for char currentChar = 'B';

answered Jan 28 '21 at 10:38

Philip Yong

1

score -1 · Answer 7 · edited May 22 '17 at 18:30

-1

The first argument of strcat must have enough space to hold the rest of the string. "" is a constant string and as such GCC does not allocate space.

Make it an array with enough space:

char buf[1024];

strcat(buf, "");
strcat(buf, "B");

edited May 22 '17 at 18:30

Peter Mortensen

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answered Jan 29 '11 at 02:53

Foo Bah

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strcat concat a char onto a string?

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