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Home/ Questions/Q 6738173
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T11:20:07+00:00 2026-05-26T11:20:07+00:00

I have the following function: int do_something(…,const float writeArray[],…){ printf(%d,*writeArray); return 0; } how

  • 0

I have the following function:

int do_something(...,const float writeArray[],...){
printf("%d",*writeArray);
return 0;
}

how I use this function is:

do_something(...,&data[0],...);

where data[0]=1

obviously my code isn’t as simple as this, but this is an illustration of the problem. When I run do_something, I get a printf value of a very large number like 23259894325. What am I doing wrong? How should I make it printf 1 like its supposed to?

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  1. Editorial Team
    Editorial Team
    2026-05-26T11:20:08+00:00Added an answer on May 26, 2026 at 11:20 am
    printf("%d",*writeArray);
    

    First off, you’re printing the first element of the array in a really weird way. If you want to do that, do writeArray[0] so other people can read your code.

    Second, %d is for print integers, but writeArray is full of floats. Try using %f. You need to use the right format string for each type.

    do_something(...,&data[0],...);
    

    This is really confusing. You want a pointer to the first element of the array? Or you’re just trying to pass the entire array? If you want to pass the array (like your later function signature suggests), just do this:

    do_something(...,data,...);
    

    Arrays turn into pointers when you pass them to functions anyway.

    If you really do want a pointer, change your second function signature to show that:

    int do_something(...,const float *writeArray,...){
    
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