Casting int to char*

Question

I have a little problem with casting int to char* (string)... is it even possible in C? I'll try to explain why i need this.

I can cast int to char but I need cast int to char*.

I had a int varriable (int number_of_revisions) and I need convert this number of revisions to char * becouse I need create a name of file and the number of revision is part of the name.... so there is part of code for better imagination of this problem.

int number_of_revision = 970; // 970 just for example
char * version;
char * new_name;
char ch_number_of_rev[4];

version = "0.";
itoa(number_of_revision,ch_number_of_rev,10);
//strcat(version, ch_num_o_rev ); // doesn't work becouse ch_number_of_rev is char and strcat requires char*

please I need quick help... Have anybody any idea how to do it? ...

score 4 · Answer 1 · edited May 23 '17 at 11:49

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but I need cast int to char*

Casting only changes the type - it does not change the value within the variable. If you need to convert an int to array of chars (i.e. a string) then use sprintf or snprintf:

char* buffer = ... allocate a buffer ...
int value = 970;
sprintf(buffer, "%d", value);

Converting int to string in c

Also, you have not allocated any memory for version - use malloc and allocate some memory.

edited May 23 '17 at 11:49

Community

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answered Nov 26 '13 at 13:53

Sadique

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I don't need change the value.. I need convert int to char* or keep it at it is itoa(number_of_revision,ch_number_of_rev,10); but then I will need join variable version which is char* with variable ch_number_of_revision wich is char.... – user3036674 Nov 26 '13 at 13:57
@user3036674 - the code i posted does what you want i believe. Test it once. – Sadique Nov 26 '13 at 13:59
Thanks I'll try the sprintf() :) – user3036674 Nov 26 '13 at 14:03
Yeah it works, thanks, but when I try to use variable "buffer" in strcat() it doesn't work: – user3036674 Nov 26 '13 at 14:08
char buffer[3]; int value = 970; sprintf(buffer, "%d", value); strcat(version, buffer ); – user3036674 Nov 26 '13 at 14:10
You have not allocated any memory for `version` - use malloc and allocate some memory. – Sadique Nov 26 '13 at 14:11
Yeah, I allocated memory for "version" later, but still doesn't work.... :( char * version =(char*) malloc (30+1); – user3036674 Nov 27 '13 at 06:46
@user3036674 - You allocated the wrong way i guess - check this code - it works: http://ideone.com/zJFI4j – Sadique Nov 27 '13 at 09:26

score 1 · Answer 2 · answered Nov 26 '13 at 13:56

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strcat here won't work because you haven't allocated any space to store the result in. Your version is probably in read-only memory anyway, so you'd get a segfault, otherwise you'll get memory corruption. So make sure to allocate enough space for it, e.g. by using

char version[10] = "0.";

You may want to read up on pointers first, though.

answered Nov 26 '13 at 13:56

creichen

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Ok I allocated memory for "version" and nothing changed.... – user3036674 Nov 27 '13 at 06:54
Ok I finaly solve this.. thanks guys for help. There is what I did: char version[40]; char buffer[4]; int value = 970; sprintf(buffer, "%d", value); strcat(version, buffer ); – user3036674 Nov 27 '13 at 07:06

Casting int to char*

2 Answers2