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Home/ Questions/Q 1055989
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T17:38:59+00:00 2026-05-16T17:38:59+00:00

In the following code,what is the meaning of buf = malloc(n * sizeof(char)); is

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In the following code,what is the meaning of buf = malloc(n * sizeof(char));

is n*sizeof(char) necessary,if yes.. please elaborate.

int n;

char* buf;

fstat(fd, &fs);

n = fs.st_size;

buf = malloc(n * sizeof(char));

EDIT1 And What if I write (n*sizeof(double))

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  1. Editorial Team
    Editorial Team
    2026-05-16T17:38:59+00:00Added an answer on May 16, 2026 at 5:38 pm

    The malloc function allocates bytes and takes as input the number of bytes you would like to allocate. The sizeof operator returns the number of bytes for a given type. In this case a char is 1 byte, but in the case of an int it is most likely 4 bytes or double is most likely 8 bytes. The expression n * sizeof(char) converts the number of char into the number of bytes that are desired.

    In the case illustrated, using char, computing the number of bytes is probably not needed, but it should be done as it helps to convey your intent.

    What the expression is doing is allocating the desired amount of memory needed to hold at most n number of char‘s and returning you a pointer, buf, to the beginning of that allocated memory.

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