allocating memory to an array of string

Question

I would like to fill an array of strings using two functions: the first, if I have n strings to allocate, will allocate n memory spaces; the second will allocate memory for each string read

Here is the first function:

char** allocate(int n)
{
    char** t;
    t=(char**)malloc(n*sizeof(char*));
    if(!t) exit(-1);
    return t;
}

Here is the second one:

void fill(char*** t,int n)
    {
        int i;
        char* help=" ";
        for(i=0;i<n;i++)
        {
            printf("\n saisir la chaine n %d :",i+1);
                scanf("%s",help);
                *t[i]=(char*)malloc((strlen(help)+1)*sizeof(char));
                strcpy(*t[i],help);
        }
    }

I did not forget to call the second one in main like this : fill(&t,n);

The problem is that I get an error after reading the first string and program ends.

To expand on that: when you call scanf("%s", help) then help must point to a memory location large enough to hold the string that is being read in. — Tim Pierce
– Tim Pierce, Commented Dec 17, 2013 at 18:59
@qwrrty Um, no. Firstly because it's not a distraction but a piece of good advice, secondly because it's not silly. — user529758
– user529758, Commented Dec 17, 2013 at 20:06

alk · Accepted Answer · 2013-12-17 19:44:33Z

2

This line

char* help=" ";

just defines a pointer pointing to " ".

There is no memory allocated to then store the data to be scanned in via scanf().

If you have a maximum of character to be scanned in do it as follows:

#define SCAN_MAXIMUM (255)
#define SCAN_FMT_STRINGIFY(max) "%"#max"s"
#define SCAN_FMT(max) SCAN_FMT_STRINGIFY(max)

...

  char help[SCAN_MAXIMUM + 1]; /* Add one for the road^H^H^H^H`0`-terminator. */
  scanf(SCAN_FMT(SCAN_MAXIMUM), help);

Also these lines do not what you want:

    *t[i]=(char*)malloc((strlen(help)+1)*sizeof(char));
    strcpy(*t[i],help);

The [] operator binds tighter then the * operator, so the lines should look like

    (*t)[i] = malloc((strlen(help) + 1));
    strcpy((*t)[i], help);

Also^2: There is no need in C to cast the results of malloc/calloc/realloc, nor is it recommnended.

Also^3: sizeof(char) is defined to be equal 1.

edited Dec 17, 2013 at 19:44

answered Dec 17, 2013 at 19:23

alk

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Comments

V-X · Accepted Answer · 2013-12-17 19:22:19Z

0

The problem is that you do not allocate the memory for help variable.

Change char* help=" "; to char help[512]="";

this way help points to a string literal (constant stored in memory block, which is not allowed to be changed.

answered Dec 17, 2013 at 19:22

V-X

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1 Comment

alk Over a year ago

@user3104033: Be aware that entering more then 511 characters will make the program fail again miserably, as the allocated memory will be overwritten and undefined behaviour will arise,

unwind · Accepted Answer · 2013-12-17 20:43:46Z

0

fill() should just take char * *, I think you're getting a operator precedence problem.

Since you don't need to re-allocate the array inside fill(), there's no point to have three levels of indirection. You should change it to just two, and of course call without the &.

edited Dec 17, 2013 at 20:43

answered Dec 17, 2013 at 19:05

unwind

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2 Comments

Amine Hosni Over a year ago

As you told me, I took char** and I changed char* help=" "; to char aide[]=" "; Now the programs asks me for all the strings, but the problem now occures after writing the last string with this error : "Stack around the variable 'aide' was corrupted"

Tim Pierce Over a year ago

If t is declared char **t and he passes the address of this variable as fill (&t, n) then the declaration char ***t was quite correct. I agree that there appears to be no need for the additional level of indirection in the code that has been posted here, though, and if t does not get re-allocated then it makes sense to remove unnecessary levels of redirection.

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