I don't see where I'm clobbering the memory

[Chad]

When the following functions takes the string "this is a string"

static void *build_string(char *s)
{
int k;
char *start;
size_t len;
size_t max;

len = strlen(s);

if(len == 0) {
fprintf(stderr,"Zero length");
exit(EXIT_FAILURE);
}

max = 1024/len;

char *p = malloc((max*len) + sizeof(p));

if(p == NULL){
fprintf(stderr, "Out of memory\n");
return NULL;
}

start = p;
for(k=0; k < max; k++){
memcpy(p, s, len);
p += len;
}

*p = '\0';
return start;
}


I get the following on my output
ou
this is a string
ou
this is a string
ou
this is a string
ou
this is a string
ou
this is a string
ou
this is a string

The ou is interlaced with my output. I figure I might be clobbering my
memory. Ideas?

Chad

 

[Chad]

When the following functions takes the string "this is a string"

static void *build_string(char *s)
{
int k;
char *start;
size_t len;
size_t max;

len = strlen(s);

if(len == 0) {
fprintf(stderr,"Zero length");
exit(EXIT_FAILURE);
}

max = 1024/len;

char *p = malloc((max*len) + sizeof(p));

if(p == NULL){
fprintf(stderr, "Out of memory\n");
return NULL;
}

start = p;
for(k=0; k < max; k++){
memcpy(p, s, len);
p += len;
}

*p = '\0';
return start;

}

I get the following on my output
ou
this is a string
ou
this is a string
ou
this is a string
ou
this is a string
ou
this is a string
ou
this is a string

The ou is interlaced with my output. I figure I might be clobbering my
memory. Ideas?

Chad

Never mind. I forgot to add a '\0' to the string s.

 

[Keith Thompson]

When the following functions takes the string "this is a string"

static void *build_string(char *s)
{
int k;
char *start;
size_t len;
size_t max;

len = strlen(s);

if(len == 0) {
fprintf(stderr,"Zero length");
exit(EXIT_FAILURE);
}

max = 1024/len;

char *p = malloc((max*len) + sizeof(p));

[snip]

You already mentioned the lack of space for the '\0', but why the
"sizeof(p)" term? Why do you need to allocate space for a pointer in
a character array?

 

[Harald van =?UTF-8?B?RMSzaw==?=]

When the following functions takes the string "this is a string"

static void *build_string(char *s)
{
int k;
char *start;
size_t len;
size_t max;

len = strlen(s);

if(len == 0) {
fprintf(stderr,"Zero length");
exit(EXIT_FAILURE);
}

max = 1024/len;

char *p = malloc((max*len) + sizeof(p));

[snip]

You already mentioned the lack of space for the '\0', but why the
"sizeof(p)" term? Why do you need to allocate space for a pointer in
a character array?

I suspect he meant sizeof(*p), for the terminating '\0'. Chad mentioned the
function argument was not terminated, but given correct input, build_string
does correctly add a '\0'.

 

[Chad]

When the following functions takes the string "this is a string"

static void *build_string(char *s)
{
int k;
char *start;
size_t len;
size_t max;

len = strlen(s);

if(len == 0) {
fprintf(stderr,"Zero length");
exit(EXIT_FAILURE);
}

max = 1024/len;

char *p = malloc((max*len) + sizeof(p));

[snip]

You already mentioned the lack of space for the '\0', but why the
"sizeof(p)" term? Why do you need to allocate space for a pointer in
a character array?

--

At the time, I was trying to allocate space for the '\0'. However,
after I stopped to think about it, I realized that this might have
been a tad bit boneheaded. So I changed

char *p = malloc((max*len) + sizeof(p));

to

char *p = malloc((max*len) + 1);

Now, I want to make a passing comment to the people that are thinking
"Oh sweet lord, this is hobbyist code." The actual build_string()
function is a bit more complex. What I posted was a stripped down
version of the actual build_string() function.

 

[Barry]

When the following functions takes the string "this is a string"

static void *build_string(char *s)
{
int k;
char *start;
size_t len;
size_t max;

len = strlen(s);

if(len == 0) {
fprintf(stderr,"Zero length");
exit(EXIT_FAILURE);
}

max = 1024/len;

char *p = malloc((max*len) + sizeof(p));

[snip]

You already mentioned the lack of space for the '\0', but why the
"sizeof(p)" term? Why do you need to allocate space for a pointer in
a character array?

--

At the time, I was trying to allocate space for the '\0'. However,
after I stopped to think about it, I realized that this might have
been a tad bit boneheaded. So I changed

char *p = malloc((max*len) + sizeof(p));

to

char *p = malloc((max*len) + 1);

Now, I want to make a passing comment to the people that are thinking
"Oh sweet lord, this is hobbyist code." The actual build_string()
function is a bit more complex. What I posted was a stripped down
version of the actual build_string() function.

Folks here reply to C questions. Providing the simple version
was the best way to get an accurate answer!

Of course responses have to assume there are not any nuances
your simple version has failed to show.

In a passing reply to you, I would suggest learning the debugger
for your set of tools.

 

[Chad]

When the following functions takes the string "this is a string"
static void *build_string(char *s)
{
int k;
char *start;
size_t len;
size_t max;
len = strlen(s);
if(len == 0) {
fprintf(stderr,"Zero length");
exit(EXIT_FAILURE);
}
max = 1024/len;
char *p = malloc((max*len) + sizeof(p));
[snip]
You already mentioned the lack of space for the '\0', but why the
"sizeof(p)" term? Why do you need to allocate space for a pointer in
a character array?
--

At the time, I was trying to allocate space for the '\0'. However,
after I stopped to think about it, I realized that this might have
been a tad bit boneheaded. So I changed

char *p = malloc((max*len) + sizeof(p));

char *p = malloc((max*len) + 1);

Now, I want to make a passing comment to the people that are thinking
"Oh sweet lord, this is hobbyist code." The actual build_string()
function is a bit more complex. What I posted was a stripped down
version of the actual build_string() function.

Folks here reply to C questions. Providing the simple version
was the best way to get an accurate answer!

Of course responses have to assume there are not any nuances
your simple version has failed to show.

In a passing reply to you, I would suggest learning the debugger
for your set of tools.

I normally use a debugger on regular basis. I just had a brain lapse
when I posted the question, Shortly after the post, I realized that I
inserted the break point in the wrong part of the code. Hence why I
didn't see one string clobbering the other string.

Chad