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Home/ Questions/Q 613281
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T17:59:08+00:00 2026-05-13T17:59:08+00:00

I got the following problem: I have an array of char pointers char *opts[]

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I got the following problem:

I have an array of char pointers

char *opts[] = { "-a", "--append", "-b" };

and a command name stored in

char cmd[] = "ls";

Now I need to compute all possible combinations, which I’ve done using the GNU Scientific Library and execute the command with the compute combinations.

The problem I have is how to compute the amount of memory I need for the char * getting passed to system().

Here’s my first try:

int length = strlen(cmd) * sizeof(char);

for (int i = 0; i < 3; ++i) {
    length += strlen(opts[i]) * sizeof(char);
}

Well, it worked, but I sense that this is not the right/best solution. Can’t I cast the two-dimensional-array to a flat one-dimensional-array and get the size of it (if I’m not wrong, the are no multidimensional arrays in C, C only mimics them).

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  1. Editorial Team
    Editorial Team
    2026-05-13T17:59:09+00:00Added an answer on May 13, 2026 at 5:59 pm

    Can’t I cast the two-dimensional-array to a flat one-dimensional-array

    Technically not true, but irrelevant here, as you do not have a two-dimensional array. You have a 1-dimensional array of values which happen to have a length. There is no guarentee that those strings are in contiguous memory, so it would be impossible to measure their total sizze in one step.

    The way to have it is the only way to get th evalue you want.

    P.S. Note that sizeof(char) is guaranteed to equal 1.

    P.P.S While I’ve already marked-down tommie75’s answer, we can put part of it to use:

    int numelem = sizeof(opts) / sizeof(opts[0]);
    for (int i = 0; i < numelum; ++i) 
    { 
        length += strlen(opts[i]) * sizeof(char); 
    } 
    
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