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Editorial Team
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Asked: 2026-05-23T20:29:44+00:00

Why does the following work and not throw some kind of segmentation fault? char

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Why does the following work and not throw some kind of segmentation fault?

char *path = "/usr/bin/";
char *random = "012";

// path + random + \0
// so its malloc(13), but I get 16 bytes due to memory alignment (im on 32bit)
newPath = (char *) malloc(strlen(path) + strlen(random) + 1);

strcat(newPath, path);
strcat(newPath, "random");
// newPath is now: "/usr/bin/012\0" which makes 13 characters.

However, if I add

strcat(newPath, "RANDOMBUNNIES");

shouldn’t this call fail, because strcat uses more memory than allocated? Consequently, shouldn’t

free(newPath)

also fail because it tries to free 16 bytes but I used 26 bytes (“/usr/bin/012RANDOMBUNNIES\0”)?

Thank you so much in advance!

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1 Answer

  1. Editorial Team

    Most often this kind of overrun problems doesn’t make your program explode in a cloud of smoke and the smell of burnt sulphur. It’s more subtle: the variable that is allocated after the overrun variable will be altered, causing unexplainable and seemingly random behavior of the program later on.

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