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Home/ Questions/Q 1063243
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T18:45:17+00:00 2026-05-16T18:45:17+00:00

in a C program I have this variable: float (*data)[3]; data = (float (*)[3])calloc(ntimes,

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in a C program I have this variable:

float (*data)[3];
data = (float (*)[3])calloc(ntimes, sizeof(float[3]));

which later is initialized. Afterwards I want to pass all the info in “data” to a serie of ntimes variables called “data_i” that I define with:

typedef float (*el_i)[3];
el_i data_i[NTIMES];  // NTIMES is a constant with value "ntimes"
for (i = 0; i < ntimes; i++) data_i[i] = (float (*)[3])calloc(1, sizeof(float[3]));

and then I try to pass the info from “data” to all the “data_i””s with

    for (i = 0; i < ntimes; i++){
      data_i[i][0][0]=data[i][0];
      data_i[i][0][1]=data[i][1];
      data_i[i][0][2]=data[i][2];
    }

Do you think here that data_i[i][0][0] is the right syntax ? Is this the better way to do this transfer or copy of information ?
Thanks

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  1. Editorial Team
    Editorial Team
    2026-05-16T18:45:18+00:00Added an answer on May 16, 2026 at 6:45 pm

    If you always calloc exactly 1 array, you can only index the array pointer with 0 anyway. What’s the reason for not taking float* then and omitting the one index step?

    typedef float *el_i;
    el_i data_i[NTIMES];
    for (i = 0; i < ntimes; i++) data_i[i] = (el_i)calloc(1, sizeof(float[3]));
    
    for (i = 0; i < ntimes; i++){
      data_i[i][0]=data[i][0];
      data_i[i][1]=data[i][1];
      data_i[i][2]=data[i][2];
    }
    

    I don’t really see the whole purpose of this.

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