The elephant in the room here is that the range of an int could be as small as -32767 to +32767, and the behaviour on assigning a larger value than this to such an int is undefined.
But, as for your main point, indeed it should concern you as it is a very bad habit. Things could go wrong as yes, 1e7 is a floating point double type.
The fact that i will be converted to a floating point due to type promotion rules is somewhat moot: the real damage is done if there is unexpected truncation of the apparent integral literal. By the way of a “proof by example”, consider first the loop
for (std::uint64_t i = std::numeric_limits<std::uint64_t>::max() - 1024; i ++< 18446744073709551615ULL; ){
std::cout << i << "\n";
}
This outputs every consecutive value of i in the range, as you’d expect. Note that std::numeric_limits<std::uint64_t>::max() is 18446744073709551615ULL, which is 1 less than the 64th power of 2. (Here I’m using a slide-like “operator” ++< which is useful when working with unsigned types. Many folk consider --> and ++< as obfuscating but in scientific programming they are common, particularly -->.)
Now on my machine, a double is an IEEE754 64 bit floating point. (Such as scheme is particularly good at representing powers of 2 exactly – IEEE754 can represent powers of 2 up to 1022 exactly.) So 18,446,744,073,709,551,616 (the 64th power of 2) can be represented exactly as a double. The nearest representable number before that is 18,446,744,073,709,550,592 (which is 1024 less).
So now let’s write the loop as
for (std::uint64_t i = std::numeric_limits<std::uint64_t>::max() - 1024; i ++< 1.8446744073709551615e19; ){
std::cout << i << "\n";
}
On my machine that will only output one value of i: 18,446,744,073,709,550,592 (the number that we’ve already seen). This proves that 1.8446744073709551615e19 is a floating point type. If the compiler was allowed to treat the literal as an integral type then the output of the two loops would be equivalent.