Haskell, 45 43 32 bytes

foldl(\n h->n*10+fromEnum h-48)0

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it's the same as this but it with foldl it's is similar to a reduce-op:

n#(h:t)=       - # -> infix function taking a number and a string(head:tail)
(n*10+fromEnum h-48)#t  
                 - compute first char and call recursively on tail

n#_=n          - end condition

f=(0#)         - apply 0 to #

Haskell, 45 43 bytes

n#(h:t)=(n*10+fromEnum h-48)#t
n#_=n
f=(0#)

Try it online!

n#(h:t)=       - # -> infix function taking a number and a string(head:tail)
(n*10+fromEnum h-48)#t  
                 - compute first char and call recursively on tail

n#_=n          - end condition

f=(0#)         - apply 0 to #

Haskell, 45 43 bytes

n#(h:t)=(n*10+fromEnum h-48)#t
n#_=n
f=(0#)

Try it online!

n#(h:t)=       - # -> infix function taking a number and a string(head:tail)
(n*10+fromEnum h-48)#t  
                 - compute first char and call recursively on tail

n#_=n          - end condition

f=(0#)         - apply 0 to #

Haskell, 45 43 32 bytes

foldl(\n h->n*10+fromEnum h-48)0

Try it online!

it's the same as this but it with foldl it's is similar to a reduce-op:

n#(h:t)=       - # -> infix function taking a number and a string(head:tail)
(n*10+fromEnum h-48)#t  
                 - compute first char and call recursively on tail

n#_=n          - end condition

f=(0#)         - apply 0 to #

Haskell, 45 bytes

n#(h:t)=(n*10+fromEnum h-48)#t
n#[]=n
f s=0#s

Try it online!

Haskell, 45 43 bytes

n#(h:t)=(n*10+fromEnum h-48)#t
n#_=n
f=(0#)

Try it online!

n#(h:t)=       - # -> infix function taking a number and a string(head:tail)
(n*10+fromEnum h-48)#t  
                 - compute first char and call recursively on tail

n#_=n          - end condition

f=(0#)         - apply 0 to #