Assign new size to const char* using memset()

Question

I have declare buffer const char* buf;

Later on I want to re assign size using memset

buffer_len = 1024;
    memset(buf, '\0', buffer_len);
    buf[strlen(buf)-1]='\0';

gives error:

client.cpp:73:30: error: invalid conversion from ‘const void*’ to ‘void*’ [-fpermissive]
In file included from client.cpp:2:0:
/usr/include/string.h:62:14: error:   initializing argument 1 of ‘void* memset(void*, int, size_t)’ [-fpermissive]
client.cpp:75:21: error: assignment of read-only location ‘*(buf + (((sizetype)strlen(buf)) + -1u))’

I know it's due to const but is there any alternative or way to perform it event it is const?

Answer 1

The assignment buf[strlen(buf)-1]='\\0'; is invalid because you defined buf as const: const char* buf; Read compiler's error message: error: assignment of read-only location .

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One point: You set buf with nul \\0 so length of buf is 0 ( \\0 at zero index) then if suppose you don't declare buf as const even then you would be assiging at negative index because strlen(buf) - 1 == 0 - 1 = -1 – Undefined behaviour

Answer 2

memset does not assign size. It fills a buffer with bytes. Filling a buffer declared as const char* makes no sense, since the reason you declare it const is for yourself not to write to it.

You could create a different array instead, since this const does not prevent you from changing the pointer itself.

Reassigning size should probably be called reallocating memory, you can use one of malloc , calloc or others to do it. Or since you've tagged this with c++, probably using the new operator would be the best idea.

Answer 3

Obvious the author is asking a solution for what he want to operate. And simply explanation of the comping error is satisfactory. I do not want to explain why the author's code is not compiling, cause the above answers explain very well about this.

I am try to give a solution if the author really want to do all the operations he meant in his code, though I do not recommend to do that in real world software project.

Use const_cast to remove the const property of the variable:

const char* buf;
...
buffer_len = 1024;
...
char *ptr = const_cast<char *>(buf);
memset(ptr, '\0', buffer_len);
ptr[strlen(buf)-1]='\0';

Answer 4

A way to achieve flexible strings in C is to use realloc :

//Get some malloc'ed strings.
char* myString = asprintf("my cool, malloc()-allocated string\n");
char* myOtherString = asprintf("Oh, and by the way: Hello World!\n");

//Make room in myString to append the other string.
size_t finalStringSize = strlen(myString) + strlen(myOtherString);
myString = realloc(myString, finalStringSize + 1);   //+1 for termination

//Concatenate the strings
strcat(myString, myOtherString);
free(myOtherString), myOtherString = NULL;

But, of course, using C++ std::strings should be less of a hassle.