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ViewsSince printf of a string works with an address in input, why printf does not accept just "argv"
int main(int argc, char **argv){
printf("argv: %s\n",argv); // does not work and prints random stuff
printf("*argv: %s\n",*argv); // works and prints ".a.out"
}
I test with:
./a.out nop
My confusion is this:
"argv" variable in the second line has the address of the first char of "./a.out".
"*argv" variable in third line is the first char of "./a.out".
So why printf("argv: %s\n",argv); to only print "./a.out" does not work?
I know that it's wrong, but I don't know why.
·
Santiago Trujillo
1 answers
Answer question0
argv is a pointer to a pointer. It means it "points" to addresses, not directly to characters.
So at address argv, other addresses are stored, not strings.
If you want to access the first string, you have to use the address of it, which is either *argv or argv[0].
I don't know if it's clear enough, don't hesitate to ask for more clarifications.
·
Santiago Trujillo
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