I have the following lines:
char *name = malloc(strsize + 1);
and
uint8_t *data;
data = (uint8_t *)name;
It is correct? It doesn't exist a chance that the pointer *name will be interpreted bad when that conversion is done?
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Otniel-Bogdan Mercea
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Possible duplicate of [Conversion between uint8 and char in C](http://stackoverflow.com/questions/35264923/conversion-between-uint8-and-char-in-c) – gsamaras Oct 08 '16 at 10:50
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Out of interest, why don't you malloc directly to uint8_t*? – Bathsheba Oct 08 '16 at 11:19
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because i need to find a HUGE exploit in a program which was wrote by someone (he says that there is an exploit) and i need to find it and i was wondering if this is. – Otniel-Bogdan Mercea Oct 08 '16 at 12:52
2 Answers
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That shouldn't be much of a problem, except that the signedness of the memory would be interpreted differently between access along data and name. In most of the practical platforms, the size of char and uint8_t in bits is the same.
tofro
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and if they are interpreted differently you may access a memory location which you don't want? – Otniel-Bogdan Mercea Oct 08 '16 at 11:09
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@user6575913 it's the signness of the data interpreted, **not** the signness of the pointer – phuclv Oct 08 '16 at 11:11
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@user6575913 Example: `data[0] = 255;` printing `(int)name[0]` will yield -1. This can be confusing, but you're not accessing any "forbidden" memory – tofro Oct 08 '16 at 11:14
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`In most of the practical platforms, sizeof char is "1",` sizeof(char) is 1 on **all** platforms, by definition. – wildplasser Oct 08 '16 at 11:23
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No, the conversion is legal. However, problems will arise when you try to print data, because you no longer have a char*. Other than that the casting is fully supported.
fornit
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